CBSE Class 12 Chemistry Biomolecules Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. Name two diseases caused due to deficiency of enzymes.
Answer:
Albinism and phenylketonuria.

Teacher's Note:
1. Albinism results from lack of the enzyme tyrosinase; phenylketonuria from lack of phenylalanine hydroxylase.

 

Question 2. Give one example of: (a) water soluble, (b) fat soluble vitamins.
Answer:
(a) Vitamin C
(b) Vitamin D

Teacher's Note:
1. Any of A, D, E, K is fat soluble; any of B group or C is water soluble.

 

Question 3. Name a protein which is insoluble in water.
Answer:
Keratin.

Teacher's Note:
1. Fibrous proteins like keratin (hair, nails) are water insoluble; globular proteins are usually soluble.

 

Question 4. Name the deficiency disease resulting from lack of Vitamin ‘A’ in the diet.
Answer:
Night blindness, Xerophthalmia.

Teacher's Note:
1. Xerophthalmia is hardening of the cornea of the eye.

 

Question 5. Mention two important functions of carbohydrates in plants.
Answer:
Major energy source, storage molecules like starch in plants.

Teacher's Note:
1. A further function: cellulose forms the cell wall of plants.

 

Question 6. Name two of the different types of RNA molecules found in cells of organisms.
Answer:
tRNA, mRNA, rRNA.

Teacher's Note:
1. Expand them in exams: messenger RNA, transfer RNA and ribosomal RNA.

 

Question 7. The deficiency of which vitamin causes the disease pernicious anaemia?
Answer:
Vitamin B₁₂.

Teacher's Note:
1. Pernicious anaemia means a fall in RBC count; do not confuse with vitamin B₆ (convulsions).

 

Question 8. During curdling of milk, what happens to sugar present in it?
Answer:
Lactose changes to lactic acid.

Teacher's Note:
1. The lactic acid formed lowers the pH and this makes the milk proteins coagulate.

 

Question 9. The two strands in DNA are not identical but complementary. Explain.
Answer:
Base pairing rule is followed; A = T and G ≡ C.

Teacher's Note:
1. The symbols show the number of H-bonds: two between A and T, three between G and C.

 

Question 10. What are monosaccharides?
Answer:
Sugars which cannot be hydrolysed to give simpler units or compounds.

Teacher's Note:
1. Examples: glucose, fructose, galactose, ribose.

 

Question 11. What is the difference between native protein and denatured protein?
Answer:
Proteins found in a biological system with unique 3D-structure and biological activity is called native protein. When native protein is subjected to physical and chemical change, protein loses its biological activity and is called denatured protein.

Teacher's Note:
1. In denaturation the 2° and 3° structures are lost but the primary structure (sequence) stays intact.

 

Question 12. Amino acids are amphoteric in nature. Explain.
Answer:
It can react with acid and base both as per the following reaction:
NH₂–CH(R)–COOH + OH⁻ → NH₂–CH(R)–COO⁻ + H₂O
NH₂–CH(R)–COOH + H⁺ → ⁺NH₃–CH(R)–COOH

Teacher's Note:
1. The –COOH group gives the acidic character and the –NH₂ group the basic character; in water they exist as zwitter ions.

 

Question 13. What is the structural feature characterising reducing sugars?
Answer:
The main structural feature of reducing sugars is the presence of an aldehyde group (–CHO) such as in glucose, mannose, galactose, etc. or the α-ketol grouping (–CO–CH₂OH) as present in fructose.

Teacher's Note:
1. In disaccharides, a free anomeric carbon (as in maltose, lactose) makes the sugar reducing.

 

Question 14. What are biomolecules ? (Raj. Board 2011)
Answer:
Complex lifeless chemical substances which not only build up the living systems (creatures) but are also responsible for their growth, maintenance and their ability to reproduce are called biomolecules. For example, carbohydrates, proteins, nucleic acids, enzymes, hormones, vitamins, etc.

Teacher's Note:
1. Remember the key point of the definition: the molecules themselves are lifeless, yet together they make up and run living systems.

 

Question 15. How is oxygen replenished in our atmosphere ? (CBSE 2007)
Answer:
Oxygen is replenished in our atmosphere through photosynthesis in plants. During photosynthesis, CO₂ of the air (produced as a result of combustion of fuels and breathing of animals and human beings) is consumed and O₂ is generated.
x CO₂ + y H₂O \( \xrightarrow[\text{Chlorophyll}]{\text{Sunlight}} \) C_x(H₂O)_y + x O₂
(C_x(H₂O)_y = Carbohydrate)

Teacher's Note:
1. Writing the equation with sunlight and chlorophyll over the arrow earns the full mark; do not forget that a carbohydrate is the other product.

 

Question 16. Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose ? (CBSE 2014)
Answer:
HOCH₂CHOHCHOHCHOHCHOHCHO (Glucose) \( \xrightarrow{\text{HI},\ \Delta} \) CH₃CH₂CH₂CH₂CH₂CH₃ (n-Hexane)
Since n-hexane has six carbons connected in a straight chain, therefore, this reaction suggests that glucose also has six carbon atoms connected in a straight chain.

Teacher's Note:
1. Prolonged heating with HI reduces every oxygen-containing group, so only the carbon skeleton survives; that is why it reveals the straight chain.

 

Question 17. Glucose on reaction with acetic anhydride gives glucose pentacetate. What does it suggest about the structure of glucose ? (CBSE Sample Paper 2017)
Answer:
Since glucose reacts with acetic anhydride to give glucose pentaacetate, therefore, this reaction suggests that glucose contains five –OH groups.
CHO–(CHOH)₄–CH₂OH + 5(CH₃CO)₂O → CHO–(CHOCOCH₃)₄–CH₂OCOCH₃ + 5CH₃COOH
Since the pentaacetate is stable, the five –OH groups are attached to different carbon atoms.

Teacher's Note:
1. Two –OH groups on the same carbon would lose water easily; the stable pentaacetate shows each –OH sits on a different carbon.

 

Question 18. Write the product formed when D-glucose reacts with conc. HNO₃. (CBSE 2018)
Answer:
Saccharic acid (glucaric acid) is formed.
CHO–(CHOH)₄–CH₂OH \( \xrightarrow{\text{conc. HNO}_3} \) HOOC–(CHOH)₄–COOH (Saccharic acid)

Teacher's Note:
1. Nitric acid oxidises both ends (–CHO and the primary –CH₂OH) to –COOH; bromine water oxidises only –CHO and gives gluconic acid.

 

Short Answer Questions (2 Marks)

Question 19. (i) Name the disease caused by deficiency of vitamin D.
(ii) Why cannot vitamin C be stored in our body?
Answer:
(i) Rickets.
(ii) Vitamin C is a water soluble vitamin which is excreted in urine and cannot be stored in our body.

Teacher's Note:
1. Vitamin D deficiency causes rickets in children and osteomalacia in adults.

 

Question 20. Name the constituents of starch and what is the difference between them?
Answer:
Amylose: A linear polymer of α-glucose, water soluble.
Amylopectin: Branched polymer of α-glucose, water insoluble.

Teacher's Note:
1. Amylose has only C1–C4 glycosidic links; amylopectin has C1–C6 links at the branch points as well.

 

Question 21. Give reasons:
(i) On electrolysis in acidic solution amino acids migrate towards cathode while in alkaline solution these migrate towards anode.
(ii) The monoamino monocarboxylic acids have two pK_a values.
Answer:
(i) NH₂–CH(R)–COO⁻ + H₂O ⇄^{OH− +}NH₃–CH(R)–COO⁻ ⇄^{H+ +}NH₃–CH(R)–COOH
In alkaline solution the anion NH₂–CH(R)–COO⁻ is formed, which migrates towards anode; the zwitter ion is neutral and does not migrate; in acidic solution the cation ⁺NH₃–CH(R)–COOH is formed, which migrates towards cathode.
(ii) Due to zwitter ion formation.

Teacher's Note:
1. The two pK_a values belong to the –COOH group and the –NH₃⁺ group of the zwitter ion.

 

Question 22. Coagulation of egg white on boiling is an example of denaturation of protein. Explain it in terms of structural changes.
Answer:
Protein albumin present in egg white gets denatured i.e., 2° & 3° structures are destroyed and 1° structure is retained.

Teacher's Note:
1. Heat breaks the hydrogen bonds holding the folded chains; the uncoiled chains then tangle and coagulate.

 

Question 23. Describe two important functions of nucleic acids.
Answer:
(i) DNA is responsible for transfer of heredity information from one generation to another.
(ii) RNA is responsible for protein synthesis.

Teacher's Note:
1. DNA can do this because of its property of replication: each strand acts as a template for a new strand.

 

Question 24. Fructose contains a keto group but still it reduces Tollens' reagent. Explain.
Answer:
Under the basic conditions of Tollens' reagent, fructose undergoes Lobry de Bruyn van Eikenstein rearrangement. As a result, fructose gives an equilibrium mixture of fructose, glucose and mannose. Since both glucose and mannose contain –CHO group, therefore, they reduce Tollens' reagent.

Teacher's Note:
1. Fructose does not reduce bromine water (acidic medium), which is used to tell glucose from fructose.

 

Question 25. Glucose forms an oxime but glucose pentaacetate does not. Explain.
Answer:
Glucose reacts with NH₂OH via the open chain form which has the free –CH = O group to form glucose oxime. Glucose pentaacetate, on the other hand, cannot be converted into the open chain form because its anomeric hydroxyl group (i.e. C₁–OH) is acetylated and hence does not form the oxime.

Teacher's Note:
1. Ring opening needs a free OH at C-1; once that OH is acetylated the ring is locked.

 

Question 26. How do epimers differ from anomers?
Answer:
Carbohydrates which differ in configuration at the glycosidic carbon (i.e. C₁ in aldoses and C₂ in ketoses) are called anomers while those which differ in configuration at any carbon other than glycosidic carbon are called epimers.
For example, α-D-glucose and β-D-glucose are anomers since they differ in configuration at C₁ (glycosidic carbon) while glucose and mannose are called epimers since they differ in configuration at C₂ (other than the glycosidic carbon C₁). In other words, glucose and mannose are C₂-epimers. Similarly, we can show that glucose and galactose are C₄-epimers since they differ in configuration only at C₄.

Teacher's Note:
1. All anomers are epimers at the anomeric carbon, but the term ‘anomer’ is used only for the C-1 (or C-2 in ketoses) difference in cyclic forms.

 

Question 27. Can the acid chloride of an α-amino acid be made by treating it with SOCl₂?
Answer:
No. Two molecules of the initially formed α-amino acid chloride react with each other to form a dipeptide acid chloride which, in turn, reacts further to form tetrapeptide acid chloride, etc.
NH₂–CHR–COOH + SOCl₂ → NH₂–CHR–COCl + SO₂ + HCl
NH₂CHRCOCl + NH₂CHRCOCl (α-Amino acid chloride, two molecules) \( \xrightarrow{-HCl} \) NH₂CHRCONHCHRCOCl (Dipeptide acid chloride) → → etc.

Teacher's Note:
1. The free –NH₂ group of one molecule attacks the reactive –COCl of another; the amino group has to be protected first.

 

Question 28. Is a diet consisting mainly of rice an adequate diet? Why or Why not?
Answer:
A diet consisting mainly of rice is not an adequate diet because it is deficient in lysine and threonine which are essential amino acids required for growth and maintenance of health and hence their deficiency has to be supplemented by other protein rich diets like pulses, etc.

Teacher's Note:
1. Pulses are rich in lysine, which is why rice and dal together give a balanced protein.

 

Question 29. What changes occur in the nature of egg proteins on boiling?
Answer:
On boiling, the soluble globular proteins present in the egg undergo coagulation to form insoluble fibrous proteins. As a result, proteins lose their biological activity and thus get denatured. Chemically, during denaturation, the secondary and tertiary structures of proteins are destroyed but primary structure remains intact.

Teacher's Note:
1. Denaturation breaks H-bonds and other weak forces, not peptide bonds – so the amino acid sequence is unchanged.

 

Question 30. What forces are responsible for the stability of α-helix? Why is it named as 3.6₁₃ helix?
Answer:
The stability of α-helix structure is due to intramolecular H-bonding between –NH– and –CO– groups of the same polypeptide chain. The α-helix is termed as 3.6₁₃ helix since each turn of the helix has approximately 3.6 amino acids and the hydrogen bonding leads to the formation of a 13-membered ring.

Teacher's Note:
1. Intramolecular H-bonds give the α-helix; intermolecular H-bonds between neighbouring chains give the β-pleated sheet.

 

Short Answer Questions (3 Marks)

Question 31. (i) What type of linkage is present in disaccharides?
(ii) Write one source and deficiency disease of vitamin B₁₂.
(iii) Write the difference between DNA and RNA. [CBSE Comptt. Delhi 2017]
Answer:
(i) Glycosidic linkage.
(ii) Source: Meat, fish, egg, curd (any one); Pernicious anaemia
(iii) DNA is a double strand while RNA is a single strand molecule (or any other correct difference)

Teacher's Note:
1. Vitamin B₁₂ comes almost only from animal foods, which is why strict vegetarians are at risk of its deficiency.

 

Question 32. (i) What type of linkage is present in proteins?
(ii) Give one example each of water soluble and fat soluble vitamins.
(iii) Draw pyranose structure of glucose. [CBSE Comptt. Delhi 2017]
Answer:
(i) Peptide linkage
(ii) Water soluble – Vit B/C, Fat soluble – Vit A/D/E/K
(iii)

Teacher's Note:
1. In the Haworth drawing of D-glucose, the CH₂OH at C-5 is up; the anomeric –OH at C-1 is down for α and up for β.

 

Question 33. (i) Which vitamin deficiency causes rickets?
(ii) Name the base that is found in nucleotide of RNA only.
(iii) Glucose on reaction with acetic anhydride gives glucose penta acetate. What does it suggest about the structure of glucose? [CBSE SQP 2017]
Answer:
(i) Vitamin D
(ii) Uracil
(iii) 5 OH groups are present

Teacher's Note:
1. Students often get confused between the bases present in RNA and DNA: A, G, C are common; U is only in RNA and T only in DNA.
2. Since glucose is stable, the five –OH groups must be on five different carbon atoms.

 

Question 34. (i) Write one reaction of D-Glucose which cannot be explained by its open chain structure.
(ii) What type of linkage is present in Nucleic acids?
(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins? [CBSE OD 2016]
Answer:
(i) Glucose does not give 2,4-DNP test in spite of having aldehyde group and does not form the hydrogen sulphite addition product with NaHSO₃.
(ii) Phosphodiester linkages are present between the 5′ and 3′ carbon atoms in nucleic acids.
(iii) Water soluble vitamin – Vitamin C/Vitamin B₁ (Thiamine) (Any one); Fat soluble vitamin – A/Vitamin D. (Any one)

Teacher's Note:
1. For part (i), add the reason: the –CHO group is tied up with the C-5 –OH as a cyclic hemiacetal, so it is not free to react.

 

Question 35. (a) Name the branched chain component of starch.
(b) Ribose in RNA and deoxyribose in DNA differ in the structure around which carbon atom?
(c) How many peptide linkages are present in a tripeptide? [CBSE SQP 2018-2019]
Answer:
(a) Amylopectin.
(b) C-2
(c) Two peptide linkages.

Teacher's Note:
1. A peptide of n amino acids has (n − 1) peptide bonds – a tripeptide has 2.

 

Question 36. Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids.
(ii) The two strands of DNA are complementary to each other.
(iii) Reaction of glucose that indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose. [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Due to the formation of zwitter ion.
(ii) The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.
(iii) CHO–(CHOH)₄–CH₂OH \( \xrightarrow{Br_2\ water} \) HOOC–(CHOH)₄–CH₂OH (Gluconic acid)
Or glucose gets oxidised to gluconic acid on reaction with mild oxidising agent like Bromine water.

Teacher's Note:
1. In part (ii), name the pairs: adenine pairs with thymine (two H-bonds) and guanine with cytosine (three H-bonds).
2. Bromine water oxidises only an aldehyde group and keeps all six carbons, which is why this reaction proves –CHO rather than a ketone.

 

Question 37. Differentiate between the following:
(i) Secondary and tertiary structure of protein
(ii) α-helix and β-pleated sheet structure of protein
(iii) Fibrous and globular protein
Answer:
(i) Secondary structure is responsible for the shape of protein α-helix and β-pleated sheets in which polypeptide chains have peptide bonds.
Tertiary structure represents overall folding of polypeptide chain and gives rise to the fibrous or globular molecular shape.
(ii) α-helix structure: The peptide chains coiled up to form right handed helix involving H-bonding (Intramolecular).
β-pleated sheets: The peptide chains lie side by side together by intermolecular hydrogen bonding.
(iii) Fibrous proteins: the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, giving a fibre-like structure; they are insoluble in water, e.g., keratin, myosin.
Globular proteins: the polypeptide chains coil around to give a spherical shape; they are soluble in water, e.g., insulin, albumin.

Teacher's Note:
1. Remember intra- vs inter-: α-helix H-bonds are within one chain, β-sheet H-bonds are between neighbouring chains.

 

Question 38. (i) Name the four bases present in DNA.
(ii) Which of them is not present in RNA?
(iii) Give the structure of a nucleotide of DNA.
Answer:
(i) Adenine, Guanine, Thymine, Cytosine.
(ii) Thymine.
(iii)

Teacher's Note:
1. In a DNA nucleotide C2′ carries only H (deoxy); drawing an –OH there makes it an RNA nucleotide – a very common slip.

 

Question 39. Glucose or sucrose are soluble in water but cyclohexane and benzene are insoluble in water. Explain.
Answer:
Glucose contains 5 –OH groups and sucrose contains eight –OH groups, because of this they form intermolecular hydrogen bonding (with water), so they are soluble in water. But benzene and cyclohexane don’t contain –OH groups hence don’t form intermolecular hydrogen bonding, so they are not soluble in water.

Teacher's Note:
1. 'Like dissolves like': polar molecules that can H-bond with water dissolve; non-polar hydrocarbons do not.

 

Long Answer Questions (5 Marks)

Question 40. Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA. [NCERT Exemplar]
Answer:
Complete hydrolysis of DNA yields a pentose sugar, phosphoric acid and nitrogen containing heterocyclic compounds called bases.
Structures:
(i) Sugar:
(ii) Phosphoric acid:
(iii) Nitrogen base: DNA contains four bases: Adenine, Guanine, Thymine and Cytosine.

A unit formed by the attachment of a base to 1′-position of sugar is called nucleoside.
When nucleoside links to phosphoric acid at 5′-position of sugar moiety, a nucleotide is formed. Nucleotides are joined together by phosphodiester linkage between 5′- and 3′-carbon atoms of the pentose sugar.

In DNA, two chains of nucleic acid coil around each other and held together by H-bonds between bases of two chains (A pairs with T and G pairs with C).

Teacher's Note:
1. In 2-deoxyribose the C-2′ carries two H atoms and no –OH – draw it carefully, it is the feature that names the sugar.
2. In the helix diagram mark the pairs A–T and G–C and the dimensions (20 Å diameter, 3.4 Å between base pairs, 34 Å per turn) to get full marks.