Biomolecules Mock Test – Class 12 Chemistry
GKaim: Measure | Improve | Achieve

Biomolecules Mock Test – Class 12 Chemistry

Progressive Test — Guest First Round

0%

Biomolecules – Progressive Test

Welcome to the Progressive Test.

Click Start Test to begin the loaded practice round.

Good luck!

1 / 50

1. Match each sucrose feature in Column I with its suitable description in Column II.

Column I Column II
P. Monosaccharide components 1. Negative before hydrolysis
Q. Glycosidic linkage 2. Glucose and fructose
R. Fehling test 3.
S. Complete hydrolysis 4. Equimolar monosaccharides

2 / 50

2. 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:

3 / 50

3. Hydrolysis of an unknown nucleic acid yields adenine, guanine, cytosine, a pentose sugar and phosphate, but neither thymine nor uracil is detected in the analysed portion. Which additional observation would be most decisive for distinguishing DNA from RNA?

4 / 50

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

5 / 50

5. A collection of nitrogenous-base molecules contains a total of heterocyclic rings. The purines satisfy , while the pyrimidines satisfy . The value of is:

6 / 50

6. A person consumes a diet lacking fresh water-soluble vitamin sources for a prolonged period. Deficiency may develop relatively quickly because many such vitamins:

7 / 50

7. A sucrose solution is treated with invertase under suitable conditions. The enzyme:

8 / 50

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

9 / 50

9. Consider the following statements.
Statement I: The lock-and-key model treats the active site as approximately complementary to the substrate before binding.
Statement II: The induced-fit model allows the active site to adjust its conformation during substrate binding.
Statement III: The induced-fit model implies that an enzyme has no substrate specificity.
The acceptable statements are:

10 / 50

10. The historical empirical pattern for many carbohydrates is completed by:

11 / 50

11. A linear double-stranded DNA segment initially contains base pairs. A double-stranded portion equal to of its original length is removed from one end. Use per base pair. Calculate the axial length and total phosphodiester-bond count of the remaining segment.

12 / 50

12. Lactose yields on complete hydrolysis:

13 / 50

13. The molecular formula and open-chain classification of fructose are:

14 / 50

14. Assertion: Glucose gives a positive Tollens test even though its predominant aqueous forms are cyclic.
Reason: Cyclic glucose remains in equilibrium with a small open-chain aldehydic fraction.

15 / 50

15. Alanylglycine and glycylalanine are different compounds mainly because:

16 / 50

16. A sample contains of identical mononucleotides, each containing one base, one pentose and one phosphate group. Complete cleavage into separate components produces a total of:

17 / 50

17. Complete hydrolysis of carbohydrate P produces three monosaccharide molecules from each molecule of P. Carbohydrate P is best classified as:

18 / 50

18. Use the Fischer-projection description below.
The carbonyl group is placed at the top of the vertical chain. At the stereogenic carbon farthest from the carbonyl group, the hydroxyl group lies on the right.
This monosaccharide belongs to:

19 / 50

19. A sample contains of identical linear polynucleotide molecules, each containing nucleotide residues. Complete hydrolysis first cleaves every phosphodiester bond, and each resulting nucleotide then undergoes two additional hydrolytic cleavages to give separate base, pentose and phosphate components. Assuming one water molecule is consumed per bond cleaved, the total amount of water consumed is:

20 / 50

20. The tertiary structure of a protein refers mainly to:

21 / 50

21. Use the structural description below.
The open-chain sugar has the hydroxyl-group arrangement of D-glucose, with the hydroxyl group on its penultimate carbon lying to the right in the Fischer representation. Ring closure occurs when the hydroxyl group on reacts with the aldehyde group on . In the cyclic form, the anomeric hydroxyl group lies opposite to the terminal group.
The cyclic sugar is:

22 / 50

22. A hormone differs from an enzyme mainly because a hormone generally:

23 / 50

23. Consider the following statements about glucose anomers.
Statement I: Both anomers have the same molecular formula.
Statement II: They differ in configuration only at the anomeric carbon.
Statement III: They are constitutional isomers with different carbon skeletons.
The acceptable statements are:

24 / 50

24. Assertion: Every biomolecule is necessarily a naturally occurring polymer.
Reason: Glucose, vitamins and many hormones are non-polymeric biomolecules.

25 / 50

25. Beriberi is most closely associated with deficiency of:

26 / 50

26. Use the arrangement described below.
A polypeptide backbone coils regularly. Hydrogen bonds form within the same chain between peptide carbonyl and amide groups, while the side chains project outward.
The arrangement is:

27 / 50

27. Consider the following statements.
Statement I: Many water-soluble vitamins require regular dietary replacement because body storage is limited.
Statement II: Fat-soluble vitamins may accumulate in the liver or adipose tissue.
Statement III: Solubility has no relationship to vitamin storage or toxicity risk.
The acceptable statements are:

28 / 50

28. The general structure of an -amino acid is:

29 / 50

29. Which statement correctly describes the opposing roles of insulin and glucagon in blood-glucose regulation?

30 / 50

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

31 / 50

31. Consider the following statements.
Statement I: Vitamin C deficiency may impair collagen-related tissue maintenance.
Statement II: Vitamin K deficiency may impair normal blood coagulation.
Statement III: Vitamin A deficiency is the principal cause of beriberi.
The acceptable statements are:

32 / 50

32. Consider the following statements about nucleic-acid hydrolysis.
Statement I: Partial hydrolysis of a polynucleotide can produce shorter polynucleotide fragments and mononucleotides.
Statement II: More complete cleavage of a nucleotide can yield phosphate, pentose sugar and a nitrogenous base.
Statement III: Hydrolysis of nucleic acids normally converts every nitrogenous base into an amino acid.
The acceptable statements are:

33 / 50

33. Which comparison between DNA and RNA is chemically correct?

34 / 50

34. 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.

35 / 50

35. A person takes very large doses of vitamins A and D for several months without medical supervision. The principal concern is that:

36 / 50

36. Fructose can be obtained together with glucose by hydrolysis of:

37 / 50

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

38 / 50

38. The correctly completed comparison is:

Option DNA RNA
A Contains ribose and uracil Contains deoxyribose and thymine
B Usually single-stranded and less stable Always double-stranded and more stable
C Usually double-stranded and contains deoxyribose Usually single-stranded and contains ribose
D Contains peptide bonds Contains glycosidic bonds only and no phosphate

39 / 50

39. Consider the following statements.
Statement I: Glucose reduces Tollens reagent.
Statement II: Glucose reduces Fehling solution.
Statement III: The reducing behaviour of glucose is connected with the open-chain aldehydic form present in equilibrium.
The acceptable statements are:

40 / 50

40. A single linear polynucleotide chain contains nucleotide residues. The combined number of --glycosidic bonds and phosphodiester bonds in the intact chain is . Nine phosphodiester bonds are then hydrolysed. Find the original nucleotide count and the number of phosphodiester bonds remaining.

41 / 50

41. Protein denaturation is best described as:

42 / 50

42. At saturating substrate concentration, an enzyme preparation initially has a maximum rate of . A treatment denatures of the enzyme molecules. The remaining preparation is then concentrated so that its enzyme concentration becomes times the value immediately after denaturation. The new maximum rate is:

43 / 50

43. Consider the following statements about cyclic fructose.
Statement I: Fructofuranose is formed by intramolecular hemiketal formation.
Statement II: Its anomeric carbon is .
Statement III: Its - and -forms can interconvert through a ring-opened form.
The acceptable statements are:

44 / 50

44. One mole of maltose yields on complete hydrolysis:

45 / 50

45. A connected acyclic glycogen fragment is formed from glucose molecules by condensation. Its molecular formula is:

46 / 50

46. The ring oxygen atom in D-fructofuranose originates from:

47 / 50

47. Match each RNA type in Column I with its principal role in Column II.

Column I Column II
P. 1. Carries amino acids and contains an anticodon region
Q. 2. Carries coding information
R. 3. Major structural and catalytic component of ribosomes
S. Ribozyme 4. RNA molecule capable of catalysis

48 / 50

48. Match each reaction of glucose in Column I with the structural conclusion in Column II.

Column I Column II
P. Oxime formation 1. Five hydroxyl groups
Q. Oxidation with bromine water 2. Carbonyl group
R. Pentaacetate formation 3. Straight six-carbon chain
S. Prolonged heating with 4. Aldehyde group

49 / 50

49. Glycogen differs from amylopectin mainly because glycogen:

50 / 50

50. Examine the following statements.
Statement I: Protein denaturation may destroy biological activity without hydrolysing the primary peptide sequence.
Statement II: An enzyme changes the equilibrium constant of the reaction it catalyses.
Statement III: Excessive intake of some fat-soluble vitamins may cause accumulation-related toxicity.
Statement IV: A circulating hormone directly affects only cells capable of recognising its signal.
The acceptable statements are:

Your score is

Share your achievement!

LinkedIn Facebook
0%

Complete 100% of the Progressive Test coverage to unlock Mistake Review.
Complete 100% of the Progressive Test coverage to unlock Certificate Challenge.

Subscribe
Notify of
guest
0 Comments
Scroll to Top