Chemistry · Volume 2 · Chapter 14

Samacheer Class 12 Chemistry - Biomolecules

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Q.1 Which one of the following rotates the plane polarized light towards left? a) D(+) Glucose b) L(+) Glucose c) D(-) Fructose d) D(+) Galactose
Answer: c

Levorotation (rotation to the left) is indicated by a negative sign. D(-)-fructose has a negative specific rotation (≈ −92°) and is therefore levorotatory. The other listed sugars are dextrorotatory.

Q.2 The correct corresponding order of names of four aldoses with configuration given below respectively is: a) L-Erythrose, L-Threose, L-Erythrose, D-Threose b) D-Threose, D-Erythrose, L-Threose, L-Erythrose c) L-Erythrose, L-Threose, D-Erythrose, D-Threose d) D-Erythrose, D-Threose, L-Erythrose, L-Threose
Answer: d

Erythrose has the two chiral centres with identical configurations (meso-like arrangement in the D/L sense) while threose has opposite configurations. Matching the conventional D/L assignment for the pictured configurations gives option (d). (Note: exact mapping depends on the given Fischer projections; answer chosen as the conventional correct sequence.)

Q.3 Which one given below is a non-reducing sugar? a) Glucose b) Sucrose c) Maltose d) Lactose
Answer: b

Sucrose is non-reducing because the glycosidic bond involves the anomeric carbons of both glucose and fructose (α-1,β-2 linkage), so no free hemiacetal/hemiketal is available to be oxidized.

Q.4 Glucose --(HCN)--> Product --(hydrolysis)--> Product --(HI + Heat)--> A. The compound A is: a) Heptanoic acid b) 2-Iodohexane c) Heptane d) Heptanol
Answer: c

Addition of HCN to the aldehydic group gives a cyanohydrin (adds one carbon). Hydrolysis of the nitrile (−CN) gives the corresponding carboxylic acid (one carbon longer than the original sugar aldehyde). Subsequent treatment with HI and heat reduces/decarboxylates to the corresponding alkane. Starting from a 6‑carbon aldose (glucose open chain), the final alkane is a 7‑carbon alkane, heptane.

Q.5 Assertion: A solution of sucrose in water is dextrorotatory. But on hydrolysis in the presence of little hydrochloric acid, it becomes levorotatory. Reason: Sucrose hydrolysis gives equal amounts of glucose and fructose. As a result of this change in sign of rotation is observed. Choose correct option.
Answer: a

Sucrose is dextrorotatory (+34°). Hydrolysis yields equimolar glucose (≈ +52.7°) and fructose (≈ −92.4°). The algebraic sum (+52.7 − 92.4 ≈ −39.7°) is levorotatory, so the reason correctly explains the assertion.

Q.6 The central dogma of molecular genetics states that the genetic information flows from: a) Amino acids → Protein → DNA b) DNA → Carbohydrates → Proteins c) DNA → RNA → Proteins d) DNA → RNA → Carbohydrates
Answer: c

The central dogma: DNA is transcribed to RNA, which is translated to protein (DNA → RNA → Protein).

Q.7 In a protein, various amino acids linked together by: a) Peptide bond b) Dative bond c) α - Glycosidic bond d) β - Glycosidic bond
Answer: a

Amino acids in proteins are joined by peptide (amide) bonds between the α-carboxyl of one amino acid and the α-amino of the next.

Q.8 Among the following the achiral amino acid is: a) 2-ethylalanine b) 2-methylglycine c) 2-hydroxymethylserine d) Tryptophan
Answer: b

2‑Methylglycine (commonly sarcosine, N‑methylglycine) has an α‑carbon that is CH2 (two identical H substituents), so it has no stereogenic centre and is achiral. The other listed amino acids have a chiral α‑carbon.

Q.9 The correct statement regarding RNA and DNA respectively is: a) RNA sugar is arabinose and DNA sugar is ribose b) RNA sugar is 2'-deoxyribose and DNA sugar is arabinose c) RNA sugar is arabinose and DNA sugar is 2'-deoxyribose d) RNA sugar is ribose and DNA sugar is 2'-deoxyribose
Answer: d

RNA contains D‑ribose (has an OH at 2′) while DNA contains 2′‑deoxy‑D‑ribose (lacking the 2′‑OH).

Q.10 In aqueous solution an amino acid mostly exists in: a) NH2-CH(R)-COOH b) NH2-CH(R)-COO- c) H3N+-CH(R)-COOH d) H3N+-CH(R)-COO-
Answer: d

Amino acids in aqueous neutral solution exist predominantly as zwitterions: H3N+–CH(R)–COO− (protonated amino group and deprotonated carboxylate).

Q.11 Which one of the following is not produced by body? a) DNA b) Enzymes c) Hormones d) Vitamins
Answer: d

Most vitamins are essential nutrients obtained from the diet (although some can be synthesized by gut flora to a limited extent); in general the human body does not synthesize all required vitamins.

Q.12 The number of sp2 and sp3 hybridised carbon in fructose are respectively: a) 1 and 4 b) 4 and 2 c) 5 and 1 d) 1 and 5
Answer: d

Open‑chain D‑fructose has one sp2 carbon (the ketone carbon C‑2) and the remaining five carbons are sp3, so sp2:sp3 = 1:5.

Q.13 Vitamin B2 is also known as: a) Riboflavin b) Thiamine c) Nicotinamide d) Pyridoxine
Answer: a

Vitamin B2 is commonly called riboflavin.

Q.14 The pyrimidine bases present in DNA are: a) Cytosine and Adenine b) Cytosine and Guanine c) Cytosine and Thiamine d) Cytosine and Uracil
Answer: c

The pyrimidine bases in DNA are cytosine and thymine (often misspelled 'thiamine' in some texts). Uracil is found in RNA, not DNA.

Q.15 Among the following, which structure corresponds to L‑serine? (structures not provided in text)
Answer: cannot determine

The question refers to structural diagrams that are not present in the provided text. L‑serine is the enantiomer with the amino group on the left in the standard Fischer projection (i.e. S configuration at the α‑carbon for serine). Without the actual structures/diagrams it is not possible to identify which labeled option corresponds to L‑serine.

Q.16 The secondary structure of a protein refers to: a) fixed configuration of the polypeptide backbone b) hydrophobic interaction c) sequence of α-amino acids d) α-helical backbone
Answer: a

Secondary structure refers to local ordered conformations of the polypeptide backbone (e.g. α‑helix, β‑sheet) stabilized mainly by backbone hydrogen bonds — i.e. fixed configurations of the backbone. Option (d) is a specific example (α‑helix).

Q.17 Which of the following vitamins is water soluble? a) Vitamin E b) Vitamin K c) Vitamin A d) Vitamin B
Answer: d

The B‑group vitamins are water‑soluble. Vitamins A, E and K are fat‑soluble.

Q.18 Complete hydrolysis of cellulose gives: a) L-Glucose b) D-Fructose c) D-Ribose d) D-Glucose
Answer: d

Cellulose is a polymer of β‑D‑glucose; complete hydrolysis yields D‑glucose.

Q.19 Which of the following statement is not correct? a) Ovalbumin is a simple food reserve in egg-white b) Blood proteins thrombin and fibrinogen are involved in blood clotting c) Denaturation makes protein more active d) Insulin maintains the sugar level in the human body.
Answer: c

Denaturation disrupts the native three‑dimensional structure of a protein, usually causing loss (not gain) of biological activity. Thus statement (c) is incorrect.

Q.20 Glucose is an aldose. Which one of the following reactions is not expected with glucose? a) It does not form oxime b) It does not react with Grignard reagent c) It does not form osazones d) It does not reduce Tollens reagent
Answer: b

Glucose, being an aldehyde (aldose), forms oximes and osazones and gives a positive Tollens test (is reducing). However, unprotected glucose has multiple acidic OH groups that destroy Grignard reagents, so under ordinary conditions glucose does not react with Grignard reagents unless the hydroxyls are protected. Therefore (b) is correct.

Q.21 If one strand of the DNA has the sequence 'ATGCTTGA', then the sequence of complementary strand would be a) TACGAACT b) TCCGAACT c) TACGTACT d) TACGRAGT
Answer: a

Base pairing: A–T and G–C. Complementary of 5'-ATGCTTGA-3' is 3'-TACGAACT-5', read 5'→3' as TACGAACT.

Q.22 Insulin, a hormone chemically is a) Fat b) Steroid c) Protein d) Carbohydrates
Answer: c

Insulin is a peptide hormone composed of amino acids (a protein); it consists of two polypeptide chains linked by disulfide bonds.

Q.23 α-D (+) Glucose and β-D (+) glucose are a) Epimers b) Anomers c) Enantiomers d) Conformational isomers
Answer: b

α and β forms differ only in configuration at the anomeric carbon (C‑1) of the cyclic sugar; such stereoisomers are called anomers.

Q.24 Which of the following are epimers a) D(+)-Glucose and D(+)-Galactose (b) D(+)-Glucose and D(+)-Mannose c) Neither (a) nor (b) (d) Both (a) and (b)
Answer: d

Epimers differ at a single stereogenic carbon. Glucose and galactose differ at C‑4; glucose and mannose differ at C‑2. Hence both pairs are epimers.

Q.25 Which of the following amino acids are achiral? a) Alanine b) Leucine c) Proline d) Glycine
Answer: d

Glycine has two hydrogen atoms on the α‑carbon (R = H) and therefore has no chiral center; it is achiral.

Short answer QuestionsShort answer Questions20 questions
Q.1What type of linkages hold together monomers of DNA?v
Solution

Nucleotides in a single DNA strand are joined by 3'→5' phosphodiester bonds between the 3'‑OH of one deoxyribose and the 5'‑phosphate of the next. Bases are attached to sugars by N‑glycosidic bonds.

Answer:

Phosphodiester linkages (and N‑glycosidic bonds between sugar and base).

Q.2Give the differences between primary and secondary structure of proteins.v
Solution

Primary structure = linear sequence of amino acids joined by peptide (amide) bonds; determines all higher structures. Secondary structure = regular local conformations (α‑helix, β‑pleated sheet, turns) stabilized mainly by hydrogen bonds between backbone C=O and N‑H groups; does not involve side‑chain interactions.

Answer:

Primary: amino acid sequence linked by peptide bonds. Secondary: local folding (α‑helix, β‑sheet) stabilized by H‑bonds between backbone C=O and N‑H.

Q.3Name the Vitamins whose deficiency cause i) rickets ii) scurvyv
Solution

Rickets is caused by deficiency of vitamin D (impaired Ca2+ absorption and bone mineralization). Scurvy is caused by deficiency of vitamin C, needed for collagen hydroxylation.

Answer:

i) Vitamin D ii) Vitamin C (ascorbic acid)

Q.4Write the Zwitter ion structure of alaninev
Solution

At physiological pH alanine exists as a zwitterion with the amino group protonated and the carboxyl deprotonated: \(\mathrm{H_3N^+\!-CH(CH_3)-COO^-}\).

Answer:

Zwitterion: H3N+–CH(CH3)–COO−

Q.5Give any three difference between DNA and RNAv
Solution

Additional differences: DNA is more stable, stores genetic information; RNA functions in transfer and expression (mRNA, tRNA, rRNA).

Answer:

1) Sugar: DNA has deoxyribose, RNA has ribose. 2) Bases: DNA uses thymine (T), RNA uses uracil (U). 3) Structure: DNA usually double‑stranded helix, RNA usually single‑stranded.

Q.6Write a short note on peptide bondv
Solution

Formation: condensation (dehydration) reaction giving –CO–NH– link. Resonance between lone pair on N and carbonyl gives partial double bond, restricting rotation and making peptide bond planar; central to protein backbone stability.

Answer:

Peptide bond is an amide linkage formed between the carboxyl of one amino acid and the amino group of another with elimination of water; it has partial C–N double‑bond character and is planar.

Q.7Give two difference between Hormones and vitaminsv
Solution

Also: hormones can be peptides, steroids or amines; vitamins are organic compounds classified as fat‑ or water‑soluble.

Answer:

1) Origin/function: Hormones are endogenous chemical messengers produced by endocrine glands to regulate physiology; vitamins are dietary micronutrients required for normal metabolism. 2) Quantity and role: Hormones act at low concentrations as signaling molecules; vitamins act mainly as coenzymes/antioxidants and are required in small dietary amounts.

Q.8Write a note on denaturation of proteinsv
Solution

Denaturation disrupts noncovalent interactions (H‑bonds, hydrophobic, ionic) but usually does not break primary peptide bonds. Example: cooking egg white causes albumin denaturation and precipitation.

Answer:

Denaturation = loss of native 3D structure (secondary/tertiary/quaternary) caused by heat, pH, solvents, detergents or heavy metals, resulting in loss of biological activity; may be reversible or irreversible.

Q.9What are reducing and non - reducing sugars?v
Solution

Test: reducing sugars give positive Fehling's/Tollens' tests. In sucrose both anomeric carbons are involved in the glycosidic bond, so no free reducing group.

Answer:

Reducing sugars have a free aldehyde or free ketose (in open chain) that can reduce mild oxidants (e.g., glucose, fructose, maltose); non‑reducing sugars lack a free anomeric hydroxyl (e.g., sucrose).

Q.10Why carbohydrates are generally optically active.v
Solution

The presence of multiple stereogenic centers (except in achiral cases) causes rotation of plane‑polarized light; e.g., glucose has several asymmetric carbons.

Answer:

Because most carbohydrates contain one or more chiral (asymmetric) carbon atoms, giving stereoisomerism and optical activity.

Q.11Classify the following into monosaccharides, oligosaccharides and polysaccharides. i) Starch ii) fructose iii) sucrose iv) lactose iv) maltosev
Solution

Disaccharides (sucrose, lactose, maltose) are classified under oligosaccharides (short chains), fructose is a simple monosaccharide, starch is a polymer of glucose (polysaccharide).

Answer:

i) Starch — polysaccharide; ii) Fructose — monosaccharide; iii) Sucrose — disaccharide (oligosaccharide); iv) Lactose — disaccharide (oligosaccharide); v) Maltose — disaccharide (oligosaccharide).

Q.12How are vitamins classifiedv
Solution

Fat‑soluble vitamins are stored in body fat and liver; water‑soluble vitamins are not stored long and must be regularly ingested.

Answer:

Vitamins are classified as fat‑soluble (A, D, E, K) and water‑soluble (vitamin C and B‑complex).

Q.13What are harmones? Give examplesv
Solution

Hormones act at low concentrations, have specific receptors on target cells and regulate growth, metabolism, reproduction and homeostasis.

Answer:

Hormones are chemical messengers secreted by endocrine glands into the blood to regulate target organs; examples: insulin (peptide), adrenaline (amine), estrogen/testosterone (steroids).

Q.14Write the structure of all possible dipeptides which can be obtained form glycine and alaninev
Solution

General peptide structure: H2N–CH(R1)–CO–NH–CH(R2)–COOH. Substituting R = H for Gly and R = CH3 for Ala gives: Gly–Gly: H2N–CH(H)–CO–NH–CH(H)–COOH; Gly–Ala: H2N–CH(H)–CO–NH–CH(CH3)–COOH; Ala–Gly: H2N–CH(CH3)–CO–NH–CH(H)–COOH; Ala–Ala: H2N–CH(CH3)–CO–NH–CH(CH3)–COOH.

Answer:

Four dipeptides: Gly–Gly, Gly–Ala, Ala–Gly, Ala–Ala.

Q.15Define enzymesv
Solution

They have active sites that bind substrates, form enzyme–substrate complexes, and increase reaction rates often by many orders of magnitude; some enzymes are RNA (ribozymes).

Answer:

Enzymes are biological catalysts (mostly proteins) that accelerate biochemical reactions without being consumed, by lowering activation energy and showing substrate specificity.

Q.16Write the structure of α-D (+) glucophyranosev
Solution

α-D-Glucopyranose is the cyclic hemiacetal of D-glucose (pyranose form). In the Haworth representation the ring oxygen is between C1 and C5. For the D-series groups that are on the right in the Fischer projection appear down in the Haworth. Thus the substituents around the ring are: C1 — OH down (α-anomer), C2 — OH down, C3 — OH up, C4 — OH down, C5 — CH2OH up. (Right-handed chair form places the anomeric OH axial down.)

Answer:

α-D-Glucopyranose: six-membered (pyranose) ring with the anomeric OH (C1) axial (down) and CH2OH at C5 up.

Q.17What are different types of RNA which are found in cellv
Solution

Common cellular RNAs and their functions: - mRNA (messenger RNA): carries genetic code from DNA to ribosomes for translation. - tRNA (transfer RNA): delivers specific amino acids to the ribosome; has anticodon for codon recognition. - rRNA (ribosomal RNA): structural and catalytic components of ribosomes. - snRNA (small nuclear RNA): involved in pre-mRNA splicing (spliceosome). - snoRNA (small nucleolar RNA): guides chemical modification of rRNA. - miRNA (microRNA) and siRNA (small interfering RNA): regulate gene expression by RNA interference. - piRNA: involved in transposon silencing in germ cells.

Answer:

Major types: messenger RNA (mRNA), transfer RNA (tRNA), ribosomal RNA (rRNA); other types: snRNA, snoRNA, miRNA, siRNA, piRNA.

Q.18Write a note on formation of α-helix .v
Solution

Formation of the α-helix: - The polypeptide backbone coils into a right-handed helix (common form). - Stabilization arises from internal hydrogen bonds between the carbonyl oxygen of residue i and the amide hydrogen of residue i+4. - Geometry: ~3.6 amino acids per turn, pitch ≈ 5.4 Å (axial rise ≈ 1.5 Å per residue). - Side chains (R groups) point outward from the helix, minimizing steric clashes and allowing specific interactions. - α-Helices are common in proteins and contribute to structural stability and function (e.g., membrane-spanning helices are often hydrophobic).

Answer:

α-Helix is a right-handed protein secondary structure stabilized by hydrogen bonds between C=O of residue i and N–H of residue i+4; it has ~3.6 residues/turn and a pitch of ~5.4 Å.

Q.19What are the functions of lipids in living organism.v
Solution

Functions of lipids: - Energy storage: triglycerides store large amounts of energy per gram and are mobilized when needed. - Structural components: phospholipids and cholesterol are key constituents of biological membranes, controlling fluidity and permeability. - Insulation and protection: subcutaneous fat provides thermal insulation and adipose tissue cushions organs. - Signalling and regulation: steroid hormones (e.g., estrogen, testosterone), prostaglandins and other lipid mediators regulate physiological processes. - Transport and vitamins: lipids are carriers for fat-soluble vitamins (A, D, E, K) and enable their absorption. - Essential fatty acids: components of membranes and precursors for signalling molecules. - Buoyancy in some aquatic organisms and electrical insulation in nerve cells (myelin).

Answer:

Lipids function as energy stores, membrane components, thermal/electrical insulators, protective padding, signalling molecules (hormones), and sources of fat-soluble vitamins and essential fatty acids.

Q.20Is the following sugar, D - sugar or L - sugar? CHO H OH H OH H OH CH2 - OHv
Solution

Interpretation: The Fischer projection reads from top (aldehyde) to bottom (CH2OH) with three internal stereocentres shown as H left, OH right at each line given (i.e. the penultimate carbon has OH on the right). The configuration of the penultimate (next-to-last) chiral carbon determines D or L. Since the OH on the penultimate carbon is on the right, this is a D-sugar.

Answer:

D-sugar