2nd PUC Chemistry Question Bank Chapter 14 Biomolecules

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Karnataka 2nd PUC Chemistry Question Bank Chapter 14 Biomolecules

Question 1.
What are carbohydrates?
Answer:
Carbohydrates are optically active polyhydroxy aldehydes or polyhydroxy ketones or the compounds which produce such units on hydrolysis.

Question 2.
What are monosaccharides? Explain.
Answer:

Carbohydrates that cannot be hydrolysed further to give simpler units are called monosaccharides.
If a monosaccharide contain aldehyde group then it is called aldose and if it contain ketonic group then it is called ketose.
If the monosaccharide contain 3,4,5 and 6 carbon atoms then it is called trioses, tetroses, pentoses and hexoses.
Example:
Glucose → Aldohexose; Fructose → Ketohexose; Ribose → Aldopentose.

Question 3.
What are oligosaccharides?
Answer:
Carbohydrates that yield two to ten monosaccharide units on hydrolysis are called oligosaccharides.

Question 4.
What is disaccharide?
Answer:
It is an oligosaccharide which on hydrolysis yield two monosaccharide units.
Example: Sucrose → Glucose + Fructose.

Question 5.
What are polysaccharides? Give examples.
Answer:
Carbohydrates which yield a large number of monosaccharide units on hydrolysis are called polysaccharides. Example: Starch, cellulose and glycogen.

Question 6.
What are reducing and non-reducing sugars? Give examples.
Answer:
Sugars which reduces Fehlings solution and Tollen’s reagent are called reducing sugars. These sugars contain free aldehyde group or ketonic group adjacent to CHOH group. Ex: Glucose, fructose, maltose, lactose.

Sugars which does not reduce Fehlings solution and Tollen’s reagent are called non-reducing sugars. Ex: Sucrose.

Question 7.
ls sucrose is a reducing sugar? Give reason.
Answer:
Surcose is not a reducing sugar Both the reducing groups of glucose and fructose in sucrose are involved in glycosidic bond formation.

Question 8.
Elucidate the structure of glucose.
Answer:
1. Its molecular formula is found to be C₆H₁₂O₆.

2. On heating with HI it forms n-hexane suggesting that all the six carbon atoms are linked in a straight chain.

3. Glucose reacts with hydroxylamine to form an oxime. This shows it contain carbonyl group.

4. In presence of bromine water, glucose undergoes oxidation forming gluconic acid. This shows carbonyl group is an aldehyde group.

5. Acetylation of glucose with acetic anhydride gives glucose penta acetate which confirms the presence of five OH groups.

6. On oxidation with strong oxidising agent like Cone. HNO₃ it gives saccharic acid. This indicate the presence of primary alcoholic group.
Based on the above facts Fischer gave open and ring structure tor- glucose.

Question 9.
How do you show that glucose contains a linear chain of 6 carbon atoms?
Answer:
On heating with HI it forms n-hexane suggesting that all the six carbon atoms are linked in a straight chain.

Question 10.
How do you show that glucose contain carbonyl group?
Answer:
When glucose reacts with NH₂OH (hydroxylamine) glucose oxime is produced. When glucose reacts with HCN glucose cyanohydrin is obtained. These reactions shows glucose contain carbonyl group.

Question 11.
What are the conclusions drawn from the following during the structure elucidation of glucose.

Answer:
(i) Glucose contain unbranched chain of six carbon atoms.
(ii) Glucose contain carbonyl group.
(iii) Glucose contain 5 hydroxyl groups.

Question 12.
Glucose on oxidation with Br₂/water gives gluconic acid. What does this reaction indicate about the structure of glucose?
Answer:
Tins indicates that the carbonyl group m glucose is aldehyde.

Question 13.
Write the Haworth structure of ∝-Glucose.
Answer:
Haworth ring structures:

Question 14.
Give Fischer’s open ring structures and Haworth’s ring structure of fructose.
Answer:

Haworth’s structure of fructose

Question 15.
Give the Haworth’s structure of Sucrose.
Answer:

∝- D- Glucopyranose + β – D – Fructofuranose

Question 16.
What is inversion of cane sugar?
Answer:
Sucrose is detrorotatory. But hydrolysis of this sugar gives levorotatory products. Hence this phenomenon is called inversion of cane sugar.

Question 17.
Write the Haworth’s structure of maltose.
Answer:

∝- D- Glucopyranose + ∝- D- Glucopyranose

Question 18.
Write the Haworth’s structure of Lactose.
Answer:

β- D- Glucopyranose + β- D- Glucopyranose

Question 19.
Explain the structure of starch.
Answer:
Starch is a storage polysaccharide in plants. It is a polymer of ∝-D-Glucose units and it consists of two components.
(a) Water soluble amylose.
(b) Water insoluble amylopectin.
Amylose is a long linear chain with 200-1000 ∝-D-Glucose units.
Amylopectin is a branched chain polymer of ∝-D-Glucose units.

Question 20.
Write a note on glycogen.
Answer:
Glycogen is a polysaccharide stored in animal body. It is present in liver, muscles and brain. It is known as animal starch. It is a branched chain polymers of ∝-D-Glucose units. When body needs glucose, enzymes break glycogen into glucose.

Question 21.
What is cellulose?
Answer:
It is a linear polymer of β- D-Glucose units present in cell wall of tin1 plants. It is insoluble m water.

Question 22.
Give four biological importance of carbohydrates.
Answer:

Carbohydrates acts as source of energy for animals and plants.
Cell wall of plants and bacteria is made up of cellulose.
D-Ribose and 2-Deoxy-ribose is present in nucleic acids.
Cellulose in the form of cotton used in cloth.

Question 23.
What is a glycosidic linkage?
Answer:
Linkage between two monosaccharide units through oxygen atom.

Question 24.
Mention two differences in the structure of starch and cellulose.
Answer:

Starch	Cellulose
1. Made up of ∝ – D glucose units.	1. Made up of β-D glucose units.
2. Has linear and branched polymeric chains.	2. It is a linear poymer

Question 25.
Give two differences between Amylose and Amylopectin.
Answer:

Amylose	Amylopectin
1. Water soluble component of the starch.	1. Water insoluble component of the starch.
2. Linear polymer of ∝ – D glucose units.	2. Branched polymer of ∝ – D Glucose units.

Question 26.
What is amino acid?
Answer:
Amino acid is a bio-organic molecule which contain basic amino (-NH₂) group and carboxylic acid (-COOH) group.

Question 27.
What is a-amino acid? Give its general structure.
Answer:
In amino acid, if amino group and carboxylic acid group are present on the same carbon atom is called ∝-amino acid.

Question 28.
Explain the classification amino acids based on the nature.
Answer:
Amino acids are classified into three týpes. Acidic, basic and neutral.
Equal number of amino and carboxyl groups makes it neutral.
Ex: Glycine, Alanine.
More number of amino than carboxyl groups makes it basic, e.g. Lysine.
More number of carboxyl groups as compared to amino groups makes it acidic, e.g. Aspartic acid.

Question 29.
Give an example for acidic amino acid.
Answer:
Aspartic acid or glutamic acid.

Question 30.
Name the amino acid which contain sulphur and benzene ring
Answer:

Sulphur containing amino acid is cysteine.
Benzene ring containing amino acid (aromatic amino acid) is tyrosine.

Question 31.
What are essential and non-essential amino acids? Give examples.
Answer:
Amino acids which are synthesized in the body are known as non-essential amino acids.
Example: Glycine, Alanine.

Amino acids which are not synthesized in the body and must be supplied through the diet are known as essential amino acids.
Example: Lysine, Histidine.

Question 32.
What are essential amino acids? Give an example.
Answer:
Amino acids which are synthesized in the body are known as non-essential amino acids.
Example: Glycine, Alanine.

Amino acids which are not synthesized in the body and must be supplied through the diet are known as essential amino acids.
Example: Lysine, Histidine.

Question 33.
What arc essential amino acid? Is glycine an essential amino acid.
Answer:
Amino acids which are synthesized in the body are known as non-essential amino acids.
Example: Glycine, Alanine.

Amino acids which are not synthesized in the body and must be supplied through the diet are known as essential amino acids.
Example: Lysine, Histidine.
Glycine is not an essential ammo acid.

Question 34.
What is zwitter ion? Give its general structure.
Answer:
In aqueous solution at certain pH amino acids exist as dipolar ions called zwitter ion.

Question 35.
Write the general structure of zwitter ion.
Answer:
In aqueous solution at certain pH amino acids exist as dipolar ions called zwitter ion.

Question 36.
Write the zwitter ion form of Glycine and Alanine amino acids.
Answer:

Question 37.
Write the zwitter ionic structure of glycine.
Answer:

Question 38.
Name naturally occuring optically inactive ∝-amino acid.
Answer:
Glycine.

Question 39.
What is isoelectric point of the amino acid?
Answer:
In aqueous amino acid solution at certain pH, amino acid does not migrate under the influence of an electric field is called isoelectric point of that amino acid.

Question 40.
What are dipeptides? Explain.
Answer:
When two amino acids combine and bind through -CO-NH- bond (amide bond or peptide linkage) are called dipeptides. When amino group of one amino acid condense with carboxylic acid group of the other, dipeptide bond i.e. -CO-NH- bond is formed. The resulting compound is called dipeptide.

Question 41.
How many amino acids and peptide bonds are present in dipeptides, Tripeptides and Tetrapeptides?
Answer:
Dipeptide contain one peptide bond and two amino acids. Tripeptides contain two peptide bonds and three amino acids. Tetrapeptides contain three peptide bonds and four amino acids,

Question 42.
What are polypeptides and proteins?
Answer:
Polypeptides are the compounds formed by the condensation of large number of amino acids through peptide bond are called polypeptides. Polypeptides having molecular mass more than ten thousand are called proteins.

Question 43.
How proteins are classified based on their molecular shape?
Answer:
Proteins are classified into two types, based on their molecular shape:
(a) Fibrous protein: It is fibre like structure formed when the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. These proteins are insoluble in water.
Example: Keratin in hair, Myosin in muscles.

(b) Globular protein: It is formed when chains of polypeptides coil around to give a spherical shape. They are soluble in water. Example: Insulin, Albumin.

Question 44.
What are fibrous proteins? Give an example.
Answer:
Proteins are classified into two types, based on their molecular shape:
(a) Fibrous protein: It is fibre like structure formed when the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. These proteins are insoluble in water.
Example: Keratin in hair, Myosin in muscles.

(b) Globular protein: It is formed when chains of polypeptides coil around to give a spherical shape. They are soluble in water. Example: Insulin, Albumin.

Question 45.
Name the protein present in hair.
Answer:
Kertain.

Question 46.
What are hormones? Give an example.
Answer:
Hormones are biochemieal messengers made of proteins. Ex: Insulin.

Question 47.
Name the hormone which maintains the constant level of glucose in blood.
Answer:
Insulin.

Question 48.
Give an example for globular proteins.
Answer:
Alubumin, Haemoglobin. Insulin.

Question 49.
Explain structure and shape of proteins.
Answer:
Structure and shape of proteins are studied in four levels:
Primary structure: Polypeptide chains of a protein has amino acids linked with each other in a specific sequence. This is called primary structure.

Secondary structure: The secondary structure of protein is the shape in which a long polypeptide chains can exist. They found to exist in two different types of structure.
(a) ∝ – helix structure
(b) β – pleated sheet structure

In ∝ – helix, right handed screw (helix) structure is formed when polypeptide chain twist with the help of hydrogen bonds.
In β – structure all peptide chains are streched out and laid side by side which are held together by hydrogen bonding.

Tertiary structure: It represents overall folding of the polypeptide chains. It gives two major shapes.
(a) Fibrous
(b) Globular

The main force which stabilize the structures of proteins are hydrogen bonds and disulphide linkages.

Quaternary structure: Proteins are composed of two or more polypeptide subunits. The spatial arrangement of these subunits with respect to each other is known as quaternary structure.

Question 50.
What is denaturation of proteins?
Answer:
When protein is subjected to change in temperature and change in pH, it loses its biological activity. This is called denaturation of proteins. During denaturation primary structure will not be affected.
Example: (1) Coagulation of egg white (2) Curdling of milk.

Question 51.
What is meant by denaturation of proteins? Which level of the structure of protein remains intact during denaturation of globular proteins?
Answer:
When protien is subjected to change in temperature and change in pH, it loses its biological activity. This is called denaturation of proteins. During denaturation primary structure will not be affected.
Example: (1) Coagulation of egg white (2) Curdling of milk.

Question 52.
What is the effect of denaturation on the structure of proteins?
Answer:
There is no change in the primary structure of protein but their is change in the secondary and tertiary structure of proteins.

Question 53.
What are vitamins?
Answer:
Organic compound required in the diet in small amounts to perform specific biological functions for normal growth and health of the organism.

Question 54.
How vitamins are classified? Explain.
Answer:
Vitamins are classified into two types:
(a) Fat soluble vitamins: These are soluble in fat and oils, but insoluble in water.
Example: A, D, E, K.
They are stored in liver and adipose tissues.

(b) Water soluble vitamins: Vitamin B, C are in this group. These are supplied regularly in the diet, because they are readily excreted in urine and cannot be stored in the body, (except B-12).

Question 55.
Give an example for fat soluble vitamin.
Answer:
Vitamin-A, D, E, K.

Question 56.
Name a vitamin that is stored in liver and adipose tissues.
Answer:
Vitamin-A, D, E, K.

Question 57.
Give some important vitamins name and their deficiency disease.
Answer:

Name of Vitamin	Deficiency disease
a. Vitamin A	Night blindness
b. Vitamin B₁	Beriberi
c. Vitamin B₂	Cheilosis
d. Vitamin B₆	Convulsions
e. Vitamin B₁₂	Pernicious anaemia
f. Vitamin C	Scurvy (bleeding of gums)
g. Vitamin D	Rickets
h. Vitamin K	Coagulation of blood

Question 58.
1. Name the water insoluble component of starch.
2. Mention one water soluble vitamin.
3. Is lysine essential or non-essential amino acid?
Answer:
1. Amylopectin
2. Vitamin B or Vitamin C
3. It is an essential amino acid.

Question 59.
Deficiency of which vitamin causes pernicious anemia.
Answer:
Vitamin B₁₂

Question 60.
Name the disease caused by the deficiency of vitamin A.
Answer:
Night blindness.

Question 61.
Which vitamin deficiency causes rickets?
Answer:
Vitamin-D

Question 62.
Deficiency of which vitamin causes the disease scurvy.
Answer:
Vitamm-C or Ascorbic acid.

Question 63.
What are nucleic acids?
Answer:
Biomolecules in nucleus of the cell responsible for transmission of heredity characters are called nucleic acids.

Question 64.
How nucleic acids are classified?
Answer:
Nucleic acids are classified into two types:

Ribose nucleic acid (RNA)
2-deoxyribose nucleic acids (DNA).

Question 65.
Name the components of nucleic acids.
Answer:
Components of nucleic acids are:

Nitrogen bases
Pentose sugar
Phosphoric acid.

Question 66.
Name the components of RNA.
Answer:
Components of RNA are:

Nitrogen bases: Adenine (A), Guanine (G), Cytosine (C), Uracil (U).
Pentose sugar: p-D-ribose
Phosphoric acid.

Question 67.
Name the components of DNA.
Answer:
Components of DNA are:

Nitrogen bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (l1).
Pentose tsugar: β-D-deoxy ribose
Phosphoric acid.

Question 68.
Name the base present only in DNA but not in RNA.
Answer:
Thymine.

Question 69.
Which is the nitrogen base present only in RNA but not in DNA?
Answer:
Uracil.

Question 70.
Name the pentose sugar present in RNA molecule.
Answer:
D – Ribose or β-D-Ribose.

Question 71.
Name the sugar present in DNA molecule.
Answer:
β-D-2-deoxyribose.

Question 72.
What is nucleoside?
Answer:
Nucleoside is bioorganic compound formed when heterocyclic base attaches with sugar unit.

Question 73.
What are nucleotides?
Answer:
A unit formed by the attachment of nucleoside with phosphate group is called nucleotide.

Question 74.
Describe primary and secondary structure of nucleic acids.
Answer:
Information regarding the sequence of nucleotides in the chain of nucleic acid is called its primary structure.

James Watson and Francis Crick gave secondary structure of DNA. It is double strand helix structure. The two chains are wound about each and held together by hydrogen bonds. In secondary structure of RNA helices are present which are single stranded.

Question 75.
Give three biological functions of DNA.
Answer:

DNA carries genetic information generation to generation.
DNA molecules undergoes self-replication during cell division.
DNA directs biosynthesis of proteins.

Question 76.
Give biological functions of RNA.
Answer:
Biosynthesis is of proteins is earned out by various RNA molecules.

Question 77.
Name the nucleic acid which is responsible for genetic information.
Answer:
DNA.

Question 78.
Name the hormone that contain Iodine.
Answer:
Thyroxine.