Biomolecules Mock Test – Class 12 Chemistry

1. Ribose and glucose are placed in different carbon-number classes because:

2. Cellulose is best described as:

3. Glycogen differs from amylopectin mainly because glycogen:

4. Use the arrangements described below.
Arrangement P consists of two polynucleotide strands running in opposite directions, with paired bases inside and sugar–phosphate backbones outside.
Arrangement Q consists of one ribonucleotide chain folded so that complementary regions within the same molecule pair with each other.
The correct identification is:

5. Negligible activity toward the mirror-image form, despite catalysis of one optical isomer of a substrate, illustrates:

6. Complete the structural comparison correctly:

7. A mutation changes an amino-acid side chain lining an enzyme’s active site from negatively charged to non-polar. The most reasonable prediction is:

8. Study the data below.

The strongest inference from the data is:

9. Two rotation–time curves are obtained at the same temperature.
Curve P begins near and decreases.
Curve Q begins near and increases.
Both curves approach .
The common final value shows that:

10. Three separate linear polynucleotide chains contain nucleotide residues in the ratio . The three chains together contain nucleotide residues. What is the sum of the total number of phosphodiester bonds in all three chains and the number of phosphodiester bonds in the shortest chain?

11. Assertion: Hydrogen bonding is important for stabilising an -helix.
Reason: The hydrogen bonds form mainly between peptide-backbone carbonyl and amide groups within the same helical segment.

12. The historical expression “hydrate of carbon” arose because many carbohydrates approximately follow the composition pattern:

13. Glucose pentaacetate does not readily form an oxime because:

14. A mixture contains lactose and sucrose with a total amount of . Complete hydrolysis produces of galactose. What are the amount of fructose formed and the mass of water consumed?

15. Match each biomolecule class in Column I with the most suitable description in Column II.

16. The carbon atom of glucopyranose:

17. A graph shows the fraction of intact nucleic-acid molecules against time in a mildly alkaline solution. Curve P remains nearly horizontal, whereas Curve Q falls rapidly. The two samples initially contain equal amounts of DNA and RNA. Which interpretation is most appropriate?

18. A preparation initially contains linear polynucleotide chains with a total of nucleotide residues and phosphodiester bonds. Eight pairs of chains are then joined end-to-end, with one new phosphodiester bond formed for each joining event. After these joining events, how many phosphodiester bonds and linear chains are present?

19. A protein variant has the same chain length as the normal protein, but one amino-acid residue has been replaced by another. The variant folds differently and shows much lower biological activity. The observation most directly demonstrates that:

20. Match each glycogen feature in Column I with its suitable consequence in Column II.

21. Two monosaccharides have the same carbon skeleton and the same configuration at three of four stereogenic centres, but they differ at the remaining centre. Their relationship is:

22. A -pleated sheet is formed when:

23. A linear double-stranded DNA segment contains base pairs. Assume that every nucleotide contains one --glycosidic bond and that each strand is uninterrupted. What is the sum of the total number of --glycosidic bonds and phosphodiester bonds in the DNA segment?

24. At saturating substrate concentration, an enzyme preparation produces product in . A second preparation contains times as much active enzyme and is tested for under otherwise identical saturating conditions. The expected product amount is:

25. An enzyme loses activity when a required metal ion is removed, regains activity when that ion is restored, and reaches a substrate-saturation plateau after restoration. The observations together show that:

26. Assertion: Hormonal secretion often decreases after the regulated physiological variable returns toward its normal range.
Reason: Negative feedback reduces the original hormonal stimulus once the corrective response has occurred.

27. Use the Haworth arrangement described below.
The group of D-glucopyranose is above the ring plane, and the hydroxyl group at is also above the ring plane.
The structure represents:

28. A standard double-stranded DNA segment has an axial length of . Use as the separation between adjacent base pairs and base pairs per turn. What is the sum of the total number of phosphodiester bonds and the approximate number of helical turns?

29. The glucose residues in cellulose are connected mainly through:

30. Consider the following statements.
Statement I: A triose contains three carbon atoms.
Statement II: A tetrose contains four carbon atoms.
Statement III: A heptose contains seven carbon atoms.
The acceptable statements are:

31. Equal amounts, each, of a linear polypeptide containing amino-acid residues per molecule and a linear double-stranded DNA fragment containing base pairs per molecule are completely hydrolysed only at their backbone linkages. The total water consumed is . What are and the total amount of amino-acid plus nucleotide monomers formed?

32. Assertion: The open-chain structural framework of glucose can have a maximum of stereoisomers.
Reason: Open-chain glucose contains only three independent stereogenic centres.

33. Maltose shows mutarotation in aqueous solution because:

34. The molecular-formula equation for reduction of glucose to sorbitol is:

35. Glucose is called a reducing sugar because it:

36. The correctly completed hormone-classification table is:

37. In the peptide

the N-terminal and C-terminal residues are, respectively:

38. Starch serves primarily as:

39. Open-chain glucose contains four independent stereogenic centres. Assuming no symmetry reduction, the maximum number of stereoisomers possible for this structural framework is:

40. The open-chain structure is best described as:

41. Amylopectin can generally release glucose residues from more chain termini simultaneously than amylose because amylopectin:

42. Read the situation below and answer the question.
An enzyme consists of two folded polypeptide subunits and requires for activity. After heating, the subunits separate and catalytic activity is lost. Analysis shows that the amino-acid sequences of both chains remain unchanged.
Which interpretation is most complete?

43. Strong oxidation of glucose with concentrated nitric acid produces saccharic acid because:

44. Assertion: DNA is described as the principal hereditary material in most organisms rather than in all biological systems.
Reason: Some viruses use RNA as their genetic material.

45. The row that correctly distinguishes primary structure from an -helical secondary structure is:

46. Freshly prepared pure -D-glucose has a specific rotation of , while pure -D-glucose has a specific rotation of . At equilibrium, the observed specific rotation is . Assuming only these two anomers contribute and their rotations combine linearly, the amounts of and anomers present in of equilibrium glucose are:

47. Two carbohydrates P and Q each yield exactly two monosaccharide units on complete hydrolysis. P reduces Tollens reagent, whereas Q does not. The most defensible conclusion is:

48. The correctly completed RNA-function table is:

49. Match each treatment in Column I with the principal product in Column II.

50. An enzyme selectively cleaves bonds between glucose residues but does not act on bonds. It would act most directly on: