CBSE Class 12 Chemistry Biomolecules Case Study Questions Set 01

Case Study 1

The most abundant biomolecules of the living system are proteins. They are polymers of α-amino acids connected to each other by peptide linkage. Amino acids contain amino (–NH₂) and carboxyl (–COOH) group. Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. They behave like salts rather than simple amines or carboxylic acids. 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.

The amino acids, which can be synthesised in the body, are known as non-essential amino acids. On the other hand, those which cannot be synthesised in the body and must be obtained through diet, are known as essential amino acids.

Question 1. The K_a and K_b values of α-amino acid are very low. Why?
Answer:
In α-amino acids the acidic group is –N⁺H₃ instead of –COOH group as in carboxylic acids and basic group is –COO⁻ instead of –NH₂ group as in amines. That is why they have low values of K_a and K_b.

Teacher's Note:
1. Link the low K_a/K_b to the zwitter ion: the groups that actually donate/accept protons are much weaker than –COOH and –NH₂.

Question 2. Draw the structure of zwitter ion.
Answer:
R–CH(NH₂)–COOH ⇄ R–CH(N⁺H₃)–COO⁻ (Zwitter ion)

Teacher's Note:
1. The proton moves from –COOH to –NH₂; show both charges (+ on N, − on O) in the zwitter ion.

Question 3. Name the alpha amino acid which is optically inactive.
Answer:
Glycine

Teacher's Note:
1. Glycine has R = H, so its α-carbon carries two identical H atoms and is not chiral.

Question 4. Name two essential amino acids.
Answer:
Valine, Leucine

Teacher's Note:
1. Other acceptable answers: lysine, isoleucine, phenylalanine, threonine, methionine, tryptophan, histidine, arginine.

Question 5. Give any two properties of amino acids.
Answer:
(i) Amino acids are water soluble.
(ii) They have high melting point.

Teacher's Note:
1. Both properties come from the dipolar salt-like (zwitter ion) structure of amino acids.

Case Study 2

Carbohydrates, proteins, nucleic acids, etc. form the basis of life and are responsible for the growth and maintenance of living systems. Therefore, they are referred to as biomolecules. Carbohydrates are widely distributed in nature. Carbohydrates are optically active polyhydroxy aldehydes or ketones or the compounds which produce such units on hydrolysis. Glucose, fructose, sucrose, starch, cellulose, etc. are some naturally occurring carbohydrates. They act as the major source of energy for animals and human beings.

Monosaccharides are the simple carbohydrates that cannot be broken further into smaller units on hydrolysis, e.g., glucose and fructose, ribose, etc. Oligosaccharides are the carbohydrates which on hydrolysis give two to ten units of monosaccharides, e.g., sucrose, maltose, raffinose, stachyose, etc. Polysaccharides are the carbohydrates which produce a large number of monosaccharide units on hydrolysis, e.g., starch, cellulose, etc.

Question 6. Which carbohydrate provides instant energy to the body?
Answer:
Glucose

Teacher's Note:
1. Glucose needs no digestion and is absorbed directly into the blood, so it gives instant energy.

Question 7. Which disaccharide is found only in animals and not in plants?
Answer:
Lactose

Teacher's Note:
1. Lactose is milk sugar; it is made of β-D-galactose and β-D-glucose.

Question 8. Which carbohydrate is the main constituent of plant cell wall?
Answer:
Cellulose

Teacher's Note:
1. Cellulose is a straight-chain polymer of β-D-glucose; starch (α-D-glucose) is the storage form instead.

Question 9. Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
Answer:
D-(+)-Glucose and D-(+)-Galactose

Teacher's Note:
1. Do not confuse with sucrose, which gives glucose and fructose.

Question 10. Which one of the following is a polysaccharide?
Starch, Maltose, Fructose, Glucose
Answer:
Starch

Teacher's Note:
1. Maltose is a disaccharide; glucose and fructose are monosaccharides.

Case Study 3

Carbohydrates are broadly classified as monosaccharides, oligosaccharides and polysaccharides. All types of carbohydrates are detected by Molisch test. Important monosaccharides are : glucose, fructose, galactose, etc. Glucose exists in two six-membered cyclic hemiacetal forms called α- and β-anomers. All monosaccharides are reducing, i.e., they reduce Tollens’ reagent, Fehling’s solution, undergo mutarotation and form osazones.

Question 11. What is Molisch test ? What is its use ?
Answer:
Molisch test is used for detection of all types of carbohydrates, i.e., monosaccharides, disaccharides and polysaccharides.
This test consists of adding Molisch reagent (1% alcoholic solution of α-naphthol) to an aqueous solution of a carbohydrate followed by conc. H₂SO₄ along the sides of the test tube. A violet ring is formed at the junction of the two layers.

Teacher's Note:
1. Add the conc. H₂SO₄ slowly along the side so that it forms a separate lower layer; the ring appears at the junction.

Question 12. Glucose forms oxime but glucose pentaacetate does not. Why so ?
Answer:
The cyclic hemiacetal form of glucose contains an OH group at C₁ which gets hydrolysed in aqueous solution to produce open chain form of glucose. Since open chain form of glucose contains a free —CHO group, it reacts with NH₂OH to form the oxime. In contrast when glucose is reacted with acetic anhydride, the OH group at C₁ along with four other OH groups at C₂, C₃, C₄ and C₆ form a pentaacetate. Since pentaacetate of glucose does not contain a free –OH group at C₁, it cannot get hydrolysed in aqueous solution to produce open chain aldehydic form and hence glucose pentaacetate does not form an oxime.

Teacher's Note:
1. The fact that the pentaacetate does not react with NH₂OH is one of the proofs of the cyclic structure of glucose.

Question 13. Glucose is an aldohexose but fructose is a ketohexose but both give the same osazone. Give reasons.
Answer:
Glucose and fructose differ in configuration only at C₁ and C₂. Since during osazone formation, configuration at C₁ and C₂ is destroyed, therefore, glucose and fructose give the same osazone.

Teacher's Note:
1. Mannose also gives the same osazone, because it differs from glucose only at C-2.

Question 14. With suitable examples, explain the difference between anomers and epimers. Write their structures.
Answer:
Anomers differ in configuration only at C₁ while epimers differ in configuration at a carbon other than C₁. Therefore, α- and β-forms of glucose which differ in configuration only at C₁ are anomers. However, glucose and galactose differ in configuration at C₄, therefore, galactose is the C₄-epimer of glucose. Structures of α- and β-form of glucose and galactose are :

Teacher's Note:
1. Compare the two pairs: the glucose anomers differ only at C-1, while glucose and galactose differ only at C-4 (OH down in glucose, up in galactose).

Case Study 4

Proteins are made up of hundreds of α-amino acids joined through peptide bonds. About 20 α-amino acids have been isolated from the hydrolysis of most of the proteins. Out of these ten α-amino acids are essential and the remaining ten are non-essential α-amino acids. All α-amino acids exist as zwitterions and each one has its own isoelectric point. Peptides are formed by the condensation of amino group of one α-amino acid with the carboxyl group of another molecule of the same or different amino acid with the elimination of a molecule of water. They are further classified as di, tri, tetra, pentapeptides, etc. according as two, three, four, five, etc. molecules of the same or different amino acids combine together.

Question 15. Glycine exists as zwitterion but p-aminobenzoic acid does not. Explain.
Answer:
In p-aminobenzoic acid, the lone pair of electrons on the NH₂ group is donated towards the benzene ring. As a result, acidic character of —COOH group and basic character of —NH₂ group decreases. Therefore, the weakly acidic —COOH group cannot transfer a H⁺ ion to the weakly basic —NH₂ group and hence p-aminobenzoic acid does not form a zwitterion.
p-H₂N–C₆H₄–COOH (p-Aminobenzoic acid) ⇏ p-H₃N⁺–C₆H₄–COO⁻ (Zwitterion) (does not occur)
However, in glycine, no such electron-withdrawing benzene ring is present. As a result, —NH₂ group is sufficiently basic and hence accepts a proton from —COOH group to form a zwitterion.
H₂N–CH₂–COOH (Glycine) ⇌ H₃N⁺–CH₂–COO⁻ (Zwitterion)

Teacher's Note:
1. Zwitterion formation needs a reasonably basic –NH₂; an aryl amine like aniline is a much weaker base than an alkyl amine.

Question 16. What are essential and non-essential α-amino acids ? Give one example of each.
Answer:
Amino acids which human body cannot synthesize are called essential amino acids and the others which human body can synthesize are called non-essential amino acids. For example, valine, phenylalanine, etc. are essential amino acids while glycine, alanine, tyrosine are non-essential amino acids.
Since essential amino acids are required for growth and maintenance of body, they must be supplied in diet.

Teacher's Note:
1. Ten amino acids are essential for humans; learn at least two examples from each group.

Question 17. What is isoelectric point ? Write the structure of glycine at pH 2.
Answer:
The pH at which there is no net migration of the amino acid under the influence of an applied electric field is called isoelectric point. Each amino acid has a characteristic isoelectric point. For example, glycine has an isoelectric point of 6.10.
At pH 2, in the acidic range, glycine exists as cation (I).
H₃N⁺–CH₂–COO⁻ (Glycine) \( \xrightarrow{\text{H}^+} \) H₃N⁺–CH₂–COOH (I)

Teacher's Note:
1. Below its isoelectric point an amino acid is a cation (moves to the cathode); above it, an anion.

Question 18. What are peptides ? Write the structure of glycylglycine.
Answer:
Peptides are formed by the condensation of amino group of one α-amino acid with the carboxyl group of another molecule of the same or different α-amino acid with the elimination of a molecule of water. For example, two molecules of glycine combine to form the dipeptide glycylglycine.
H₂N–CH₂–CO–OH (Glycine) + H₂N–CH₂–COOH (Glycine) \( \xrightarrow{-\text{H}_2\text{O}} \) H₂N–CH₂–CO–NH–CH₂–COOH (Glycylglycine, a dipeptide; –CO–NH– = peptide bond)

Teacher's Note:
1. Glycylglycine has one peptide bond; a free –NH₂ at one end and a free –COOH at the other.

Case Study 5

Polynucleotides are called nucleic acids. Each nucleotide is made up of three parts, i.e., a pentose sugar, a heterocyclic nitrogenous base and phosphoric acid. Depending upon the nature of sugar whether, ribose or 2-deoxyribose, nucleic acids are called RNA and DNA respectively. In all, there are five nitrogenous bases, two of which are purines while the remaining three are pyrimidines. Out of these five bases, each type of nucleic acid has four of them. Whereas DNA has two strands while RNA has a single strand. The two strands in DNA are complementary and are held together by hydrogen bonds.

Question 19. What purine and pyrimidine bases are present in DNA and RNA ?
Answer:
Purines : Adenine and guanine are present in both DNA and RNA.
Pyrimidines : Thymine and cytosine are present in DNA while uracil and cytosine are present in RNA.

Teacher's Note:
1. Purines have two fused rings, pyrimidines one ring; a purine always pairs with a pyrimidine.

Question 20. What is the difference between a nucleoside and a nucleotide ?
Answer:
A nucleoside consists of a nitrogenous base (purine or pyrimidine) attached to C-1′ of a pentose sugar (ribose or deoxyribose) by a β-linkage, i.e., Base–Sugar, e.g., adenosine.
A nucleotide is formed when the C-5′ –OH of the sugar of a nucleoside is esterified with phosphoric acid, i.e., Base–Sugar–Phosphate, e.g., adenosine monophosphate (AMP). Thus a nucleotide contains all the three components of nucleic acids while a nucleoside lacks the phosphate group.

Teacher's Note:
1. Nucleotide = nucleoside + phosphate.

Question 21. The two strands of DNA are not identical but are complementary. Explain.
Answer:
In the double helix of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases : adenine always pairs with thymine (two H-bonds) and guanine always pairs with cytosine (three H-bonds). Therefore, the sequence of bases in one strand automatically decides the sequence of bases in the other strand. Hence, the two strands are not identical but complementary to each other. For example, if one strand has the sequence A-T-G-C, the other strand has T-A-C-G.

Teacher's Note:
1. Specific base pairing is also what makes accurate DNA replication possible.

Question 22. What is the functional difference between DNA and RNA ? Explain.
Answer:
The main functions of DNA and RNA are :
(i) DNA is responsible for transmission of hereditary effects from one generation to another due to unique property of replication during cell divison as a result of which two identical strands are transferred to the daughter cell.
(ii) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually, the proteins are synthesized by various RNA molecules (r-RNA, m-RNA and t-RNA) in the cell but message for the synthesis of a particular protein is given by DNA molecules.

Teacher's Note:
1. One-line summary: DNA stores and passes on genetic information; RNA uses it to make proteins.

Case Study 6

Monosaccharides can be either aldoses or ketoses. Whereas glucose and galactose are aldoses, fructose is a ketose. All monosaccharides are reducing, i.e., they reduce Tollens’ reagent and Fehling’s solution, undergo mutarotation and form osazones. However, glucose does not give some of the characteristic reactions of aldehydes.

Case Study 7

Proteins are broadly classified as globular proteins and fibrous proteins. Whereas globular proteins are water soluble, fibrous proteins are not. All enzymes are globular proteins. The complete structure of proteins is discussed at four different levels, i.e., primary, secondary, tertiary and quaternary. When a biologically active protein is subjected to change of temperature, pH, etc., it loses its biological activity and such a protein is called denatured protein.