Bihar Board - Class 12 Chemistry -Chapter 14: Biomolecules Long Answer Question

The long question answers for Chapter 14: Biomolecules of Class 12 Chemistry are provided here in English. These questions are based on the NCERT curriculum for Class 12. They are designed to help students gain a deeper understanding of the important concepts covered in the chapter. These long question answers will be useful not only for the Class 12 board exams but also for entrance exams like NEET. The questions and answers for Bihar Board Class 12 Chemistry, Chapter 14: Biomolecules, have been prepared by expert teachers at Vidyakul.  

                                  Long Question Answer 

Question.1.  Explain what is meant by
(i) a peptide linkage
(ii) a glycosidic linkage.
Answer:
(i) Peptide linkage: A peptide linkage is an amide linkage formed between the -COOH group of one a-amino acid and NH2 group of the other a-amino acid by loss of a molecule of water.

(ii) Glycosidic linkage: The two monosaccharide units are joined together through an ethereal or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through an oxygen atom is called glycosidic linkage. 

Question.2.  Name two water soluble vitamins, their sources and the diseases caused due to their deficiency in diet.
Answer: 

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Question.3.  Explain the following terms: 

(i) Invert sugar 

(ii) Polypeptides 

Answer: 

(i) Invert sugar: An equimolar mixture of glucose and fructose obtained by hydrolysis of sucrose in presence of an acid such as dil. HCl or the enzyme invertase or sucrase is called invert sugar.

(ii) Polypeptides: They are formed when several molecules of a-amino acids are joined together by peptide bonds. 

Question.4. What are reducing sugars?

Answer:  Reducing sugars are those which can act as reducing agents. They contain in them a reducing group which may be an aldehydic (-CHO) or ketogenic (>C=O) group. The characteristic reactions of reducing sugars are with Tollen’s reagent and Fehling solution. Non-reducing sugars do not give these reactions. For example, glucose, fructose, lactose etc. are reducing sugars. Sucrose is regarded as a non-reducing sugar because both glucose and fructose are linked through their aldehyde and ketonic groups by glycosidic linkage. Since these groups are not free, sucrose is a non-reducing sugar.

Question.5. Write two main functions of carbohydrates in plants.

Answer: Two major functions of carbohydrates in plants are following

a)Structural material for plant cell walls: The polysaccharide cellulose acts as the chief structural material of the plant cell walls.

(b)Reserve food material: The polysaccharide starch is the major reserve food material in the plants. It is stored in seeds and acts as the reserve food material for the tiny plant till it is capable of making its own food by photosynthesis.

Question.6. Define the following as related to proteins:

(i) Peptide linkage

(ii) Primary structure

(iii) Denaturation

Answer: (i) Peptide bond: Proteins are condensation polymers of α-amino acids in which the same or different α-amino acids are joined by peptide bonds. Chemically, a peptide bond is an amide linkage formed between -COOH group of one α-amino acid and -NH- , group of the other -amino acid by lo;ss of a molecule of water. For example,

(ii) Primary structure: Proteins may contain one or more polypeptide chains. Each . the polypeptide chain has a large number of α-amino acids which are linked to one another in a specific manner. The specific sequence in which the various amino acids present in a protein linked to one another is called its primary structure. Any change in the sequence of  -Amino acids create a different protein.

(iii) Denaturation: Each protein in the biological system has a unique three-dimensional structure and has specific biological activity. This is called native form of a protein. When a protein in its native form is subjected to a physical change such as change in temperature or a chemical change like change in pH, etc., hydrogen bonds get broken. As a result, soluble forms of proteins such as globular proteins undergo coagulation or precipitation to give fibrous proteins which are insoluble in water. This coagulation also results in loss of biological activity of the proteins and this loss in biological activity is called denaturation. During denaturation, 2and 3structures of proteins are destroyed but 1structure remains intact.

The most common example of denaturation of proteins is the coagulation of albumin present in the white of an egg. When the egg is boiled hard, the soluble globular protein present in it is denatured and is converted into insoluble fibrous protein.

Question.7. What are the common types of secondary structure 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 :

• -helix structure

• -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 polypeptide 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 - flat sheet structure.

(a) -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 -the helix are broken. This tends to increase the length of
the helix is like a spring. On releasing the tension, the hydrogen bonds are reformed, giving back the original helical shape.

(b) -Flat sheet or -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 neighboring 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 - 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 fibroin in silk fiber has a - pleated sheet structure. The characteristic mechanical properties of silk can easily be explained on the basis of its - 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.8. Differentiate between globular and fibrous proteins.
Answer: (i) Fibrous proteins: These proteins consist of linear thread like molecules which tend to lie side by side (parallel) to form fibers. The polypeptide chains in them are held together usually at many points by hydrogen bonds and some disulphide bonds. As a result,intermolecular forces of attraction are very strong and hence fibrous proteins are insoluble in water. Further, these proteins are stable to moderate changes in temperature and pH . Fibrous proteins serve as the chief structural material of animal tissues.For example, keratin in skin, hair, nails and wool, collagen in tendons, fibrosis in silk and myosin in muscles.
(ii) Globular proteins: The polypeptide chain in these proteins is folded around itself in such a way so as to give the entire protein molecule an almost spheroidal shape. The folding takes place in such a manner that hydrophobic (non-polar) parts are pushed inwards and hydrophilic (polar) parts are pushed outwards. As a result, water molecules interact strongly with the polar groups and hence globular proteins are water soluble. As compared to fibrous proteins, these are very sensitive to small changes of temperature and pH . This class of proteins include all enzymes, many hormones such as insulin from pancreas, thyroglobulin from thyroid gland, etc.

Question.9.State clearly what are known as nucleosides and nucleotides.
Answer:

 Nucleoside: A nucleoside contains only two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-The position of the purine moiety is linked to C1 of the sugar (ribose or deoxyribose) by a -linkage.

Nucleotides: A nucleotide contains all the three basic components of nucleic acids, i.e. a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by esterification of C5, -OH of the sugar of the nucleoside with phosphoric acid.

Question 10.  Describe what you understand by primary structure and secondary structure of proteins.
Answer:

1. Primary structure of proteins: Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence which is known as the primary structure of protein. 

2. Secondary structure of proteins: The conformation which the polypeptide chains assume as a result of hydrogen bonding is called the secondary structure of the protein. 

Depending upon the size of the R groups, the two different secondary structures are possible which are: 

1. -Helix structure: Intramolecular H-bonds present between the C=O of one amino acid and N-H of four amino acids. 

2. -Pleated sheet structure: The two neighboring polypeptide chains are held together by intermolecular bonds. 

Question.11.  Write the main structural difference between DNA and RNA. Of the four bases, name those which are common to both DNA and RNA.
Answer:

Question.12.  Write down the structures and names of the products formed when D-glucose is treated with 

(i) Hydroxylamine 

(ii) Acetic anhydride.
Answer:
(i) D-glucose reacts with hydroxylamine to form oxime.

(ii) D-glucose reacts with acetic anhydride to give penta-acetate.

Question.13.  Answer the following questions:
(i) Why are vitamin A and vitamin C essential for us?
(ii) What is the difference between a nucleoside and a nucleotide?
Answer:
(i) Because deficiency of vitamin A and vitamin C causes night blindness and scurvy respectively.

(ii)

Question.14.  Amino acids may be acidic, alkaline or neutral. How does this happen? What are essential and non- essential amino acids? Name one of each type.
Answer:
1. Amino acids can be broadly classified into three classes i.e. acidic, alkaline and neutral amino acids depending on the number of -NH2 group and -COOH group.

2. Acidic amino acids: Those a-amino acids such as aspartic acid, asparagine and glutamic acid which contain NO-COOH  groups and one -NH2  the group is called acidic amino acids. 

3. Alkaline or Basic amino acids: Those a-amino acids such as lysine, arginine and histidine which contain two H2 groups and one -COOH  the group is called basic amino acids.

4. Neutral amino acids: Those a-amino acids such as glycine, alanine, valine etc. which contain one -NH2 and the -COOH the group is called neutral amino acids. 

Question.15.  Differentiate between fibrous proteins and globular proteins. What is meant by the denaturation of a protein?
Answer: 

Denaturation of protein: Due to coagulation globular protein under the influence of change in temperature, change in pH etc., the native shape of protein is destroyed and biological activity is lost and formed protein is called denatured proteins and phenomenon is denaturation. 

Question.16.  (a) Write the structural and functional differences between DNA and RNA
(b) Name two components of starch.
Answer:
(a) Structural difference DNA:

Structural difference RNA:

Functional difference: DNA's main function is to control cell activities like telling each organ what to make and what to do. RNA's main function is to make protein.

(b) Components of starch: amylose and amylopectin. 

Question.17. 

(i) Deficiency of which vitamin causes scurvy? 

(ii) What type of linkage is responsible for the nation of proteins? 

(iii) Write the product formed when glucose is treated with HI. 

Answer:

(i) Vitamin C causes scurvy. 

(ii) Peptide linkages are responsible for the formation of proteins. 

(iii) Glucose is treated with HI: 

Question.18.
(i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
(ii) Why Vitamin C cannot be stored in our body?
(iii) What is the difference between a nucleoside and nucleotide?
Answer:
(i) On hydrolysis, lactose gives -D-galactose and  -D-glucose.

(ii) Vitamin C is mainly ascorbic acid which is water soluble and is readily excreted through urine and thus cannot be stored in the body.

(iii) Nucleoside: A nucleoside contains only two basic components of nucleic acids, i.e., a pentose sugar and a nitrogenous base. It is formed by the attachment of a base to l' position of sugar. Nucleoside = Sugar + Base

Nucleotides: A nucleotide contains all the three basic components of nucleic acids, i.e., a phosphoric acid group, a pentose sugar and nitrogenous base. These are formed by the esterification of C5-OH of the sugar of the nucleoside with phosphoric acid.
Nucleotide = Sugar + Base + Phosphoric acid
Nucleotide = Sugar + Base + Phosphoric acid

Question.19.
1. Write the structural difference between starch and cellulose.
2. What type of linkage is present in Nucleic acids?
3. Give one example each for fibrous protein and globular protein.
Answer:

1. Starch contains the -D-glucose as its monomer units while cellulose contains -D- glucose as its monomer units. Acids 

2. Phosphodiester linkages are present in Nucleic 

3. Globular protein: All enzymes and hormones like insulin.  Fibrous protein: Keratin in skin.

Question.20.
(a) What type of linkage is present in disaccha- ides?
(b) Write one source and deficiency disease of vitamin B12.
(c) Write the difference between DNA and RNA.
Answer:
(a) Glycosidic linkage is present in disaccharides.
(b) Eggs are the source of Vitamin B12 and its deficiency causes pernicious anemia.
(c) DNA is a double strand while RNA is a single strand molecule. 

Question.21.
(a) What type of linkage is present in proteins?
(b) Give one example of water soluble and fat soluble vitamins.
(c) Draw pyranose structure of glucose.
Answer:
(a) Peptide linkage is present in proteins.
(b) Vitamin C  is water soluble and Vitamin D is a fat soluble vitamin.
(c) Pyranose structure of glucose

Question.22. What are monosaccharides ?

Answer: Monosaccharides are carbohydrates Which cannot be hydrolysed to smaller molecules.Their general formula is (CH2O)n Where n=3-7 These are of two types: Those which contain an aldehyde group (-CHO) are called aldoses and those which contain a keto (C=O) the group is called ketoses.

They are further classified as trioses , tetroses ,pentoses , hexoses and heptoses as they contain 3 , 4 , 5 , 6 and 7 carbon atoms respectively.For example.

Question.23. The two strands in DNA are not identical but are complementary. Explain.

Answer: The two strands in a DNA molecule are held together by hydrogen bonds between purine base of one strand and the pyrimidine base of the other and vice versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds, (i.e.,C=G) and between A (adenine) and T (thiamine) through two H-bonds (i.e., A=T). Due to this base -pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complementary and not identical.

Question 24.
Define the following terms:
1. Glycosidic linkage
2. Invert sugar
3. Oligosaccharides
Answer:

1. Glycosidic linkage: The two monosaccharide is joined together through an ethereal or oxide linkage by loss of a molecule of water. Such a linkage between o monosaccharide units through oxygen atoms is called glycosidic linkage. 

2. Invert sugar: An equimolar mixture of glucose and fructose obtained by hydrolysis of sucrose in presence of and such as dil. HCl or the enzyme invertase or sucrase is led by invert sugar. 

3. Oligosaccharides: Those carbohydrates which on hydrolysis give 2-10molecules of monosaccharides are oligosaccharides.  Example: sucrose, maltose.