CBSE Class 12 Chemistry Biomolecules Assignment

Question. The correct statement in respect of protein haemoglobin is that it:
a. Acts as an oxygen carrier in the blood
b. Forms antibodies and offers resistance to diseases
c. Functions as a catalyst for biological reactions
d. Maintains blood sugar level
Answer : A

Question. The alcohol obtained by the hydrolysis of oils and fats is:
a. Glycol
b. Glycerol
c. Propanol
d. Pentanol
Answer : B

Question. Oleic, stearic and palmitic acids are:
a. Fatty acid
b. Amino acid
c. Nucleic acid
d. Essential acid
Answer : A

Question. The waxes are long chain compounds of fatty acids, which belong to the class of?
a. Esters
b. Ethers
c. Alcohols
d. Acetic acid
Answer : A

Question. Hydrolytic reaction of fats, with caustic soda, is known as:
a. Acetylation
b. Carboxylation
c. Saponification
d. Esterification
Answer : C

Question. An example for a saturated fatty acid, present in nature is:
a. Oleic acid
b. linoleic acid
c. Linolenic acid
d. Palmitic acid
Answer : D

Question. The 'acid value' of an oil or fat is measured in terms of weight of:
a. NH₄OH
b. NaOH
c. KOH
d. CH₃COOH
Answer : C

Question. Hardening of oils is caused by:
a. H₂
b. N₂
c. O₂
d. CO₂
Answer : A

Question. The base present in DNA, but not in RNA is:
a. Guanine
b. Adenine
c. Uracil
d. Thymine
Answer : D

Question. Identify the correct statement regarding enzymes:
a. Enzymes are specific biological catalysts that cannot be poisoned
b. Enzymes are normally heterogeneous catalysts that are very specific in their action
c. Enzymes are specific biological catalysts that can normally function at very high temperature (T ~ 1000K)
d. Enzymes are specific biological catalysts that possess well-defined active sites
Answer : D

Question. The energy change produced by the combustion of food is called the 'calorific value'. The highest calorific value is given by:
a. Proteins
b. Fats
c. Carbohydrates
d. Vitamins
Answer : B

Question. A nucleotide consists of:
a. Base and sugar
b. Base and phosphate
c. Sugar and phosphate
d. Base, sugar and phosphate
Answer : D

Question. The harmone that helps in the conversion of glucose to glycogen in:
a. Adrenaline
b. Insulin
c. Cortisone
d. Bile acids
Answer : B

Question. Which of the following is not a sex hormone?
a. Testosterone
b. Estrone
c. Estradiol
d. Cortisone
Answer : A

Question. The change in optical rotation, with time, of freshly prepared solution of sugar is known as:
a. Rotatory motion
b. Inversion
c. Specific rotation
d. Mutarotation
Answer : D

Question. Mutation of DNA occurs due to changes in the sequence of one of the following:
a. Bases
b. Ribose units
c. Phosphate units
d. Sugar units
Answer : A

Question. Vitamin B₁ is:
a. Riboflavin
b. Cobalamin
c. Thiamine
d. Pyridoxine
Answer : C

Question. An alternation in the base sequence of nucleic acid molecule is called:
a. Replication
b. Mutation
c. Duplication
d. Dislocation
Answer : B

Question. Vitamin B₆ is known as:
a. Pyridoxin
b. Thiamine
c. Tocopherol
d. Riboflavin
Answer : A

Question. Sugar can be tested in urine by:
a. Molisch test
b. Dunstan's test
c. Benedict's test
d. Legal's test
Answer : C

Question. A gene is a segment of a molecule of:
a. DNA
b. m-RNA
c. t-RNA
d. Protein
Answer : D

Question. Amide group is present in:
a. Lipids
b. Carbohydrates
c. Amino acids
d. Proteins
Answer : D

Question. An example of a disaccharide made up of two units of the same monosaccharides is:
a. Sucrose
b. Maltose
c. Lactose
d. None of these
Answer : B

Question. Glucose gives silver mirror with Tollen's reagent. It shows the presence of:
a. An acidic group
b. An alcoholic group
c. A ketonic group
d. An aldehydic group
Answer : D

Question. A certain compound gives negative test with ninhydrin and positive test with Benedict's solution. The compound is:
a. A protein
b. A monosaccharide
c. A lipid
d. An amino acid
Answer : B

Question. Which of the following does not show any reducing test of aldehyde?
a. Sucrose
b. Fructose
c. Maltose
d. Lactose
Answer : A

Question. The aqueous solution of a carbohydrate gives dark blue colour with iodine. It is:
a. Glucose
b. Fructose
c. Sucrose
d. Starch
Answer : D

Question. The ultimate product of oxidation of most of hydrogen and carbon in foodstuffs are:
a. H₂O alone
b. CO₂ alone
c. H₂O and CO₂O
d. None of these
Answer : C

Question. Raffinose is:
a. Trisaccharide
b. Monosaccharide
c. Disaccharide
d. None of these
Answer : A

Question. To become a carbohydrate a compound must contain at least:
a. 2 carbons
b. 3 carbons
c. 4 carbons
d. 6 carbons
Answer : B

Question. The intermediate compound formed in the conversion of starch to glucose is:
a. Lactose
b. Sucrose
c. Maltose
d. Fructose
Answer : C

Question. Glucose and mannose are:
a. Epimers
b. Anomers
c. Ketohexoses
d. Disaccharides
Answer : A

Question. A carbohydrate that cannot be hydrolysed to simpler forms is called
a. Disaccharide
b. Monosaccharide
c. Polysaccharide
d. Trisaccharide
Answer : B

Question. On heating glucose with Fehling's solution we get a precipitate whose colour is:
a. Yellow
b. Red
c. Black
d. White
Answer : B

Question. The charring of sugar, when treated with conc. H₂SO₄, is due to?
a. Oxidation
b. Reduction
c. Dehydration
d. Hydrolysis
Answer : C

Question. Optical activity is shown by
a. Glucose
b. Fructose
c. Sucrose
d. All of these
Answer : D

Question. Which of the following is the sweetest sugar?
a. Glucose
b. Fructose
c. Lactose
d. Sucrose
Answer : B

Question. Which is false?
a. Glucose is a disaccharide
b. Starch is a polysaccharide
c. Glucose and fructose are not anomers
d. Invert sugar consists of glucose and fructose
Answer : A

Question. General formula for carbohydrates is
a. C_nH_2nO_2n+2
b. C_x(H₂O)_2x
c. C_x(H₂O)_y
d. None of these
Answer : C

Question. The disaccharide present in milk is:
a. Maltose
b. Lactose
c. Sucrose
d. Cellobiose
Answer : B

Level One Questions

Question 1. Name the sugar present in milk.
Answer: Lactose.
In simple words: Milk contains a natural sugar called lactose, which is a disaccharide.

Exam Tip: Remember that milk sugar is lactose, composed of glucose and galactose units.

Question 2. How many monosaccharide units are present in it?
Answer: Two monosaccharide units are present.
In simple words: Lactose breaks down into two simple sugars upon hydrolysis.

Exam Tip: Disaccharides always yield two monosaccharide units on hydrolysis.

Question 3. What are such oligosaccharides called?
Answer: Such oligosaccharides are called disaccharides.
In simple words: Carbohydrates yielding two monosaccharide units are termed disaccharides.

Exam Tip: Oligosaccharides that yield two monosaccharide units on hydrolysis are specifically called disaccharides.

Question 4. How do you explain the presence of all the six carbon atoms in glucose in a straight chain?
Answer: On prolonged heating with HI, glucose gives \(n\)-hexane.
In simple words: Reacting glucose with hydrogen iodide forms normal hexane, proving all six carbons are linked straightly.

Exam Tip: Mention the reagent \(\text{HI}\) and the product \(n\)-hexane to score full marks.

Question 5. Name the linkage connecting monosaccharide units in polysaccharides.
Answer: Glycosidic linkage.
In simple words: Monosaccharide units are joined together by an oxide linkage known as a glycosidic bond.

Exam Tip: Glycosidic linkage is formed by the loss of a water molecule between two hydroxyl groups.

Question 6. Under what conditions glucose is converted to gluconic and saccharic acid?
Answer: Glucose is converted to gluconic acid by bromine water and to saccharic acid by conc. \(\text{HNO}_3\).
Answer: Bromine water oxidizes only the aldehyde group, whereas concentrated nitric acid oxidizes both terminal groups.
In simple words: Mild oxidizing agents give gluconic acid, while strong oxidizing agents give saccharic acid.

Exam Tip: Distinguish clearly between the mild oxidizing action of \(\text{Br}_2\) water and strong \(\text{HNO}_3\).

Question 7. Which sugar is called invert sugar?
Answer: Sucrose.
In simple words: Sucrose is called invert sugar because hydrolysis inverts its optical rotation from dextrorotatory to laevorotatory.

Question 8. During curdling of milk, what happens to sugar present in it?
Answer: It converts into lactic acid.
In simple words: Bacteria ferment lactose sugar into lactic acid during milk curdling.

Exam Tip: Lactose undergoes fermentation by microorganisms to produce sour lactic acid.

Question 9. Monosaccharide contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule are also considered for classification. In which class of monosaccharide will you place fructose?
Answer: Fructose is a ketohexose.
In simple words: Fructose contains six carbon atoms and a keto group, making it a ketohexose.

Exam Tip: Always specify both the functional group and the number of carbon atoms when classifying sugars.

Question 10. The letters 'D' or 'L' before the name of a stereoisomer of a compound indicate the correlation of configuration of that particular stereoisomer. This refers to their relation with one of the isomers of glyceraldehyde. Predict whether the following compound has 'D' or 'L' configuration.
Answer: 'L' configuration.
In simple words: Since the hydroxyl group at the reference carbon is on the left side, it has an L-configuration.

Exam Tip: Look at the lowest asymmetric carbon atom; if \(-\text{OH}\) is on the left, it is L; if on the right, it is D.

Question 11. What are constituents of Starch?
Answer: Amylose and Amylopectin.
In simple words: Starch is made of two polymer fractions: a water-soluble linear amylose and a water-insoluble branched amylopectin.

Exam Tip: Starch = Amylose (linear, 15-20%) + Amylopectin (branched, 80-85%).

Question 12. What DNA & RNA Stand for?
Answer: Deoxyribonucleic acid and Ribonucleic acid.
In simple words: DNA and RNA are nucleic acids responsible for genetic storage and protein synthesis.

Exam Tip: Spell the terms correctly without spelling errors in nucleic acid names.

Question 13. What are Zwitter ions?
Answer: A Zwitter ion is a dipolar ion formed by neutralization of acidic and basic centers present within the molecule.
In simple words: Amino acids form internal salts where the proton from the carboxyl group transfers to the amino group.

Exam Tip: Zwitterions possess both positive and negative charges, making them neutral overall.

Question 14. What is non-reducing sugar? Give example.
Answer: The groups like \(-\text{CHO}\), \(-\text{C}=\text{O}\), which are not freely available in the molecule do not answer Tollens or Fehlings test are called non-reducing sugar. E.g., sucrose.
In simple words: Sugars that cannot reduce Tollens or Fehling reagents due to bonded carbonyl groups are non-reducing.

Exam Tip: Disaccharides like sucrose have glycosidic bonds involving reducing centers, making them non-reducing.

Question 15. Define mutarotation? Give example.
Answer: The anomers of glucose i.e., alpha and beta are having specific rotation of \(+111^\circ\text{C}\) and \(+19.2^\circ\text{C}\) respectively. The mixtures of these two have a rotation of \(+52.4^\circ\text{C}\). This is called mutarotation.
In simple words: Mutarotation is the spontaneous change in optical rotation of an aqueous solution of a sugar to an equilibrium value.

Exam Tip: Note that optical rotation values in older texts are expressed with degree units.

Question 16. Amino acids are amphoteric in behavior? Explain.
Answer: They form zwitterion (dipolar ion) and behave as neutral molecules at pH 7 (isoelectric point).
In simple words: Because they have both acidic and basic groups, amino acids can react with both acids and bases.

Exam Tip: Amphoteric nature arises from the dual acidic and basic functional groups in amino acids.

Level Two Questions

Question 1. Define native state and denaturation of protein. What happens when: a. Protein is cooled to zero degree C? b. Protein is heated to \(80^\circ\text{C}\)?
Answer: Protein in the native state has definite configuration and biological activity. The higher structure of protein is affected without disturbing the primary structure is called denaturation.
a. No change.
b. The coagulation of the protein takes place.
In simple words: Native state is the natural active form of a protein, while denaturation alters its 3D structure with heat or chemicals.

Exam Tip: Primary structure remains intact during denaturation, but secondary and tertiary structures are destroyed.

Question 2. Which forces are responsible for stability of alpha Helix of protein? Why it is called \(3.6_{13}\) helix?
Answer: H-bonding. It has 3.6 amino acids in one single turn, and a 13-member ring is formed by H bonding.
In simple words: Hydrogen bonds hold the helical coil together, with 3.6 amino acid residues per turn.

Exam Tip: Memorize that \(3.6_{13}\) signifies 3.6 amino acids per turn and a 13-atom hydrogen-bonded ring.

Question 3. What are essential amino acids? Give example and what happens when it is polymerized?
Answer: Amino acids required by the body and cannot be synthesized in our body are called essential amino acids. e.g., Lysine. When it is polymerized, polypeptide chains are formed.
In simple words: Essential amino acids must be obtained from diet, and their polymerization creates proteins.

Exam Tip: Essential amino acids cannot be synthesized by human metabolic pathways.

Question 4. Glucose and sucrose are soluble in water but Cyclohexane and benzene are not soluble. Why?
Answer: Glucose and sucrose form H-bonding with water.
In simple words: Polar hydroxyl groups in sugars form strong hydrogen bonds with polar water molecules.

Exam Tip: Like dissolves like: polar compounds dissolve in polar solvents like water.

Question 5. (i) Write the sequence of base on mRNA molecule synthesized on the following strand of DNA: TATCTACCTGGA
(ii) Name a powerful antioxidant which is a water soluble vitamin.
Answer:
(i) AUAGAUGGACCU
(ii) Vitamin C
In simple words: During transcription, adenine pairs with uracil, thymine with adenine, cytosine with guanine, and guanine with cytosine.

Exam Tip: In RNA, uracil replaces thymine when pairing with adenine.

Question 6. (i) 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 a physical change like change in temperature or a chemical change like, change in pH, denaturation of protein takes place. Explain the cause.
(ii) Structures of glycine and alanine are given below. Show the peptide linkage in glycylalanine.
\(\text{H}_2\text{N}-\text{CH}_2-\text{COOH}\) (Glycine)
\(\text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{COOH}\) (Alanine)
Answer:
(i) Due to physical or chemical change, hydrogen bonds in proteins are disturbed, globules unfold and helix gets uncoiled therefore protein loses its biological activity. This is called denaturation of proteins.
(ii) \(\text{H}_2\text{N}-\text{CH}_2-\text{COOH} + \text{H}_2\text{N}-\text{CH}(\text{CH}_3)-\text{COOH} \xrightarrow{-\text{H}_2\text{O}} \text{H}_2\text{N}-\text{CH}_2-\text{CO}-\text{NH}-\text{CH}(\text{CH}_3)-\text{COOH}\) (Glycylalanine)
In simple words: Denaturation breaks stabilizing bonds, while peptide formation links amino acids by removing water.

Exam Tip: Draw the peptide bond clearly as \(-\text{CO}-\text{NH}-\).

Question 7. (i) What are the expected products of hydrolysis of lactose?
(ii) How do you explain the absence of aldehyde group in the pentaacetate of D-glucose?
Answer:
(i) It is composed of \(\beta\)-D-galactose and \(\beta\)-D-glucose. The linkage is between C1 of galactose and C4 of glucose.
(ii) The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free \(-\text{CHO}\) group.
In simple words: Hydrolysis yields galactose and glucose, and acetylation blocks the reactive carbonyl center.

Exam Tip: Acetylation involves reaction with acetic anhydride to form pentaacetate.

Question 8. (i) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
(ii) How will you distinguish \(1^\circ\) and \(2^\circ\) hydroxyl groups present in glucose?
Answer:
(i) Complete hydrolysis of DNA yields a pentose sugar, phosphoric acid and thymine.
(ii) On oxidation with nitric acid, glucose as well as gluconic acid both yield a dicarboxylic acid, saccharic acid. This indicates the presence of a primary alcoholic (\(-\text{OH}\)) group in glucose.
In simple words: Nucleotide hydrolysis yields sugar, phosphate, and base, while selective oxidation identifies alcohol types.

Exam Tip: Primary alcohol in glucose is located at carbon-6.

Question 9. Write the reactions of D-glucose which can't be explained by its open-chain structure. How can cyclic structure of glucose explain these reactions?
Answer: The following reactions and facts could not be explained by this structure:
- Despite having the aldehyde group, glucose does not give Schiff's test and it does not form the hydrogensulphite addition product with \(\text{NaHSO}_3\).
- The pentaacetate of glucose does not react with hydroxylamine indicating the absence of free \(-\text{CHO}\) group.
- Glucose is found to exist in two different crystalline forms which are named as \(\alpha\) and \(\beta\). The \(\alpha\)-form of glucose (m.p. 419 K) is obtained by crystallisation from concentrated solution of glucose at 303 K while the \(\beta\)-form (m.p. 423 K) is obtained by crystallisation from hot and saturated aqueous solution at 371 K.
In simple words: Cyclic hemiacetal formation locks the aldehyde carbon, explaining anomalous reactions and anomers.

Exam Tip: Mention both chemical anomalies and anomeric melting points for full credit.

Level Three Questions

Question 1. Write the evidences for the following on the basis open chain structure of Glucose: (i) all the six carbon atoms are linked in a straight chain. (ii) the presence of a carbonyl group (\(>\text{C}=\text{O}\)) in glucose. (iii) five \(-\text{OH}\) groups are attached to different carbon atoms.
Answer:
(i) \(\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{HI}, \Delta} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3\) (\(n\)-Hexane)
(ii) \(\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{NH}_2\text{OH}} \text{CH}=\text{N}-\text{OH}\) and with \(\text{HCN}\) gives cyanohydrin.
(iii) \(\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH} \xrightarrow{\text{Acetic anhydride}} \text{CHO}-(\text{CH}-\text{O}-\text{CO}-\text{CH}_3)_4-\text{CH}_2-\text{O}-\text{CO}-\text{CH}_3 \text{ (Pentaacetate)}\)
In simple words: HI gives straight-chain hexane, hydroxylamine/HCN proves carbonyl presence, and acetic anhydride shows five hydroxyl groups.

Exam Tip: Clearly write out the chemical reagents and corresponding products for each reaction test.

Question 2. Explain the terms primary structure of proteins.
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 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 amino acids creates a different protein.
In simple words: The exact sequence of amino acids in a polypeptide chain is its primary structure.

Exam Tip: Primary structure determines all higher-order protein structures and biological functions.

Question 3. Explain the terms secondary structure of proteins. What is the difference between \(\alpha\)-helix and \(\beta\)-pleated sheet 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. \(\alpha\)-helix and \(\beta\)-pleated sheet structure. These structures arise due to the regular folding of the backbone of the polypeptide chain due to hydrogen bonding between \(-\text{CO}-\) and \(-\text{NH}-\) groups of the peptide bond. \(\alpha\)-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 \(-\text{NH}\) group of each amino acid residue hydrogen bonded to the \(\text{C}=\text{O}\) of an adjacent turn of the helix.
In \(\beta\)-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 known as \(\beta\)-pleated sheet.
In simple words: Secondary structures are local folded shapes stabilized by hydrogen bonds between backbone atoms.

Exam Tip: Contrast coiled \(\alpha\)-helix with extended sheet-like \(\beta\)-pleated structures.

Question 4. Explain tertiary structure of Protein.
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 \(2^\circ\) and \(3^\circ\) structures of proteins are hydrogen bonds, disulphide linkages, van der Waals and electrostatic forces of attraction.
In simple words: Tertiary structure is the comprehensive 3D folding pattern driven by side-chain interactions.

Exam Tip: Mention stabilizing forces like disulfide bridges and hydrogen bonds.

Question 5. What are structural difference between Cellulose and Starch?
Answer: Starch is a polymer of \(\alpha\)-glucose and consists of two components - Amylose and Amylopectin. Amylose is water soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 \(\alpha\)-D-(+)-glucose units held by \(\text{C}1-\text{C}4\) glycosidic linkage.
Cellulose is a straight chain polysaccharide composed only of \(\beta\)-D-glucose units which are joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.
In simple words: Starch uses \(\alpha\)-glucose with branched structures, while cellulose uses \(\beta\)-glucose in rigid straight chains.

Exam Tip: Highlight the difference between \(\alpha\)-glucose units in starch and \(\beta\)-glucose units in cellulose.