Biomolecules Mock Test – Class 12 Chemistry

1. Match each water-soluble vitamin feature in Column I with its suitable description in Column II.

2. Swollen or bleeding gums, poor wound healing and weakening of connective tissue most strongly suggest deficiency of:

3. The structure that directly represents a polyhydroxy aldehyde is:

4. A protein preparation contains amino-acid residues distributed among separate linear chains and has peptide bonds. If of its peptide bonds are hydrolysed, what are the number of peptide fragments formed and the number of peptide bonds remaining?

5. 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:

6. The name “aldohexose” communicates that the monosaccharide:

7. 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?

8. After a meal, person P shows a rise in blood glucose followed by increased glucose uptake and glycogen formation. During an overnight fast, person Q shows mobilisation of liver glycogen and release of glucose into blood. The hormones acting most directly in P and Q, respectively, are:

9. A dietary deficiency leaves several enzyme proteins present but inactive because an organic helper cannot be formed. In the same patient, a glandular messenger still reaches target cells but produces no response because the receptors are defective. The two failures most directly involve:

10. In glucose pentaacetate, the prefix “penta” is directly related to the number of ______ groups present in glucose before acetylation.

11. A preparation contains amino-acid residues distributed among separate linear polypeptide chains. If no cyclic chains or interchain peptide bonds are present, the total number of peptide bonds is:

12. A sample contains glucose, vitamin C, glycine and starch. The macromolecule in this set is:

13. Complete the structural comparison correctly:

14. Consider the following statements.
Statement I: Fructose reduces Tollens reagent in alkaline medium.
Statement II: This reaction proves that the normal open-chain structure of fructose is an aldose.
Statement III: Enediol rearrangement can produce glucose- and mannose-related aldose forms.
The acceptable statements are:

15. A linear double-stranded DNA molecule contains base pairs. Four complete double-strand breaks occur at different positions, and no nucleotide is lost. After breakage, what are the total number of phosphodiester bonds remaining and the total number of termini?

16. Two peptides contain the same amino acids in the same proportions but arrange them in different orders. The peptides differ in:

17. Match each category in Column I with the most suitable description in Column II.

18. The label D in D-glucose indicates that:

19. Pure -D-glucose has an initial specific rotation of approximately , but its aqueous solution eventually reaches approximately . This decrease occurs because:

20. Consider the following statements about nucleic acids.
Statement I: DNA and RNA are polymers of nucleotides.
Statement II: Nucleic acids commonly contain carbon, hydrogen, oxygen, nitrogen and phosphorus.
Statement III: Every nucleic acid is constructed from amino-acid residues.
The acceptable statements are:

21. Residues that are widely separated in the primary sequence become neighbours in an enzyme’s active site. This arrangement is produced mainly by:

22. Consider the following statements about amylopectin.
Statement I: Its main chains contain linkages.
Statement II: Its branch points contain linkages.
Statement III: It is completely unbranched.
The acceptable statements are:

23. Assertion: Reduction of fructose gives a mixture containing sorbitol and mannitol.
Reason: Reduction of the planar carbonyl group at can produce two configurations at the newly formed alcohol-bearing carbon.

24. A child can see normally in bright daylight but has increasing difficulty seeing after entering a dimly lit room. The most likely nutritional deficiency is:

25. Assertion: Complementary base pairing enables a nucleic-acid strand to guide formation of another strand with a related sequence.
Reason: Each base shows selective pairing with a suitable complementary base.

26. Match each structural feature in Column I with its suitable consequence in Column II.

27. Removal of the phosphate group from a mononucleotide produces:

28. Assertion: Glycogen can generally be mobilised more rapidly than a comparable unbranched glucose polymer.
Reason: Its frequent branches create many non-reducing ends at which enzymes can act simultaneously.

29. A mixture contains nucleosides and mononucleotides. Of all nitrogenous bases present, are purines. Complete cleavage separates every nucleoside into base and sugar and every nucleotide into base, sugar and phosphate. Determine the total amounts of separated component particles and nitrogenous rings.

30. Three biomolecules are described below.
Molecule P is a polymer whose sequence stores hereditary information.
Molecule Q is a small non-polymeric substance that assists biochemical regulation.
Molecule R is a polymer of amino-acid residues and may act as a catalyst.
The identities of P, Q and R, respectively, are:

31. Complete hydrolysis of a mixed nucleic-acid sample yields nitrogenous bases. Purines constitute of the bases, with . Among the pyrimidines, . The total number of guanine and uracil bases is:

32. Glycine is achiral because its -carbon:

33. Starch and cellulose are both polymers of glucose, yet they differ greatly in structure and digestibility. The strongest explanation is that:

34. Protein denaturation is best described as:

35. A lactose sample is treated with excess Tollens reagent before and after complete hydrolysis. Assume each mole of reducing sugar deposits silver. The masses of silver deposited before and after hydrolysis are, respectively, :

36. The vitamin that contains cobalt as part of its molecular structure is:

37. 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:

38. Partial hydrolysis of an unknown tripeptide gives the dipeptides Ala–Gly and Gly–Val. Independent end-group analysis shows alanine at the N terminus and valine at the C terminus. The tripeptide sequence is:

39. Assertion: Both oxime formation and cyanohydrin formation support the existence of a carbonyl-containing form of glucose.
Reason: Hydroxylamine and hydrogen cyanide can react by addition or condensation at a carbonyl carbon.

40. Assertion: Epimers and anomers both differ in configuration at one carbon atom.
Reason: Anomers differ specifically at the anomeric carbon formed during ring formation, whereas ordinary epimers differ at another stereogenic centre.

41. Assertion: Glucose and fructose form the same osazone but give different products on reduction.
Reason: Osazone formation removes differences involving and , whereas reduction preserves the consequences of their different carbonyl positions.

42. Humans obtain little glucose directly from dietary cellulose because humans:

43. A graph plots target-cell response against hormone concentration. The response rises steeply at low concentration and then approaches a plateau at high concentration. The most suitable interpretation is:

44. Match each observation in Column I with its most suitable interpretation in Column II.

45. The open-chain structure of glucose is considered incomplete mainly because it cannot explain why glucose:

46. Regular exposure of skin to suitable sunlight can contribute to:

47. The product obtained when only the aldehydic end of glucose is oxidised is:

48. The mirror image of a D-series monosaccharide, provided every stereogenic centre is inverted, is:

49. Assertion: Mutarotation requires temporary formation of open-chain glucose.
Reason: Direct conversion between the two anomers requires loss and re-formation of configuration at the planar carbonyl carbon.

50. In -D-glucopyranose, the anomeric hydroxyl group and the group are: