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. 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.

2 / 50

2. A laboratory record gives the following observations.

Sample Observation after treatment Most direct interpretation
P Peptide sequence unchanged; helices disrupted Secondary structure affected
Q One chain loses its overall fold; peptide bonds intact Tertiary structure affected
R Folded subunits separate from one another Quaternary structure affected

The interpretation of the record is:

3 / 50

3. A graph shows amounts of three species during stepwise hydrolysis of a polynucleotide. Curve P decreases continuously from the start. Curve Q initially rises, reaches a maximum and then decreases. Curve R begins near zero and rises steadily toward a final plateau. Which interpretation is most appropriate?

4 / 50

4. Protein denaturation is best described as:

5 / 50

5. Assertion: The - and -forms of D-glucopyranose are anomers.
Reason: They differ only in configuration at the hemiacetal carbon .

6 / 50

6. In a Haworth representation of a D-fructofuranose, the anomeric hydroxyl group at lies on the same side of the ring as the reference group. The structure is:

7 / 50

7. Read the situation below and answer the question.
Complete hydrolysis of a macromolecule yields deoxyribose, phosphate and the bases adenine, thymine, guanine and cytosine. The intact double-stranded molecule has an axial length of . Adenine constitutes of all its nucleotides. Use per base pair.
Which identification and composition are correct?

8 / 50

8. 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:

9 / 50

9. The correctly completed row relating a structural level to its main stabilising feature is:

Option Protein structural level Main stabilising feature
A Primary Hydrogen bonds between neighbouring strands
B Secondary Association of several folded subunits
C Primary Covalent peptide bonds joining amino-acid residues
D Quaternary Sequence of residues within one chain only

10 / 50

10. 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?

11 / 50

11. Use the arrangement described below.
Two extended polypeptide segments run in opposite N-to-C directions. Backbone hydrogen bonds connect the neighbouring segments, and side chains alternate above and below the sheet.
This structure is:

12 / 50

12. 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.

13 / 50

13. 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:

14 / 50

14. Evaluate the following statements about -pleated sheets.
Statement I: Their backbone segments may be parallel or antiparallel.
Statement II: Their side chains commonly alternate above and below the sheet.
Statement III: They are stabilised only by covalent peptide bonds between neighbouring strands.
The acceptable statements are:

15 / 50

15. A graph shows blood-glucose concentration against time after equal glucose intake by two individuals. Curve P rises and returns close to its initial value relatively quickly. Curve Q rises higher and remains elevated for much longer. Which interpretation is most appropriate?

16 / 50

16. Fructose gives positive Tollens and Fehling tests in alkaline medium even though it is a ketose because it:

17 / 50

17. During protein synthesis, molecule P carries a sequence copied from DNA, molecule Q carries an amino acid and recognises a three-base coding unit, and molecule R forms a major part of the ribosome. The correct identification is:

18 / 50

18. Four unknown biological substances give the observations below.

Substance Observation
P Hydrolysis yields amino acids; heating destroys activity without releasing free amino acids
Q Hydrolysis yields ribose, phosphate and nitrogenous bases including uracil
R Hydrolysis yields only glucose; the polymer contains linkages
S Required in small amounts; prolonged excessive intake may cause accumulation in adipose tissue

The correct identification is:

19 / 50

19. 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?

20 / 50

20. The row in which the disaccharide, constituent monosaccharides, glycosidic linkage and reducing behaviour are all correctly stated is:

Option Disaccharide Constituent units Glycosidic linkage Behaviour
A Lactose Galactose and glucose Reducing
B Maltose Two glucose units Reducing
C Sucrose Glucose and fructose Reducing
D Lactose Galactose and glucose Reducing

21 / 50

21. Hypervitaminosis is more strongly associated with prolonged excessive intake of fat-soluble vitamins because:

22 / 50

22. Match each fructose feature in Column I with its suitable description in Column II.

Column I Column II
P. 1. Anomeric carbon
Q. Oxygen of the hydroxyl group 2. Ring oxygen
R. Fructofuranose 3. Five-membered cyclic hemiketal
S. - and -forms 4. Differ in configuration at

23 / 50

23. A mixture contains -D-glucose and -D-glucose. If their specific rotations are and , respectively, the approximate specific rotation of the mixture is:

24 / 50

24. 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.

25 / 50

25. Use the arrangement described below.
An open-chain monosaccharide contains seven carbon atoms. Its carbonyl carbon is , bonded to carbon groups on both sides.
The molecule is classified as:

26 / 50

26. Consider the following statements.
Statement I: Hydrolysability and reducing behaviour are separate carbohydrate properties.
Statement II: Every hydrolysable disaccharide is reducing before hydrolysis.
Statement III: A sweet carbohydrate may be either reducing or non-reducing.
The acceptable statements are:

27 / 50

27. A colourless disaccharide solution gives a negative Tollens test. After heating with dilute acid, the resulting solution gives a positive Tollens test and contains equal amounts of glucose and fructose. The original disaccharide is:

28 / 50

28. Assertion: Condensation between suitable biomolecular units commonly eliminates a small molecule such as water.
Reason: Hydrolysis cleaves a linkage by consuming water.

29 / 50

29. Use the arrangement described below.
The anomeric hydroxyl group at of an -D-glucose unit reacts with the hydroxyl group at of another glucose unit.
The linkage formed is:

30 / 50

30. An open-chain monosaccharide has six carbon atoms and its carbonyl group is located at . It is classified as:

31 / 50

31. In the zwitterionic form of an amino acid, the principal charged groups are:

32 / 50

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

33 / 50

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

34 / 50

34. Use the arrangement described below.
A DNA template segment runs from to . A new RNA strand is assembled beside it by complementary pairing and grows from to .
Which statement correctly describes the arrangement?

35 / 50

35. A sample contains of mononucleotides, of which contain purine bases. Each purine has two fused rings, while each pyrimidine has one ring. After complete cleavage, every mononucleotide gives one base, one pentose and one phosphate unit. How many moles of nitrogenous rings and separated component particles are obtained, respectively?

36 / 50

36. A double-stranded DNA segment contains base pairs and a total of hydrogen bonds. Using per base pair, what are the number of pairs and the axial length of the segment?

37 / 50

37. Examine the following claims.
Claim I: Glucose and fructose are functional isomers.
Claim II: Both can reduce Fehling solution, but their reasons for doing so are not identical.
Claim III: Both produce only one alditol on reduction.
The acceptable claims are:

38 / 50

38. Amino acids are amphoteric because they can:

39 / 50

39. The isoelectric point, , of an amino acid is the at which the amino acid:

40 / 50

40. For an equilibrium mixture containing only and anomers, the observed specific rotation is correctly represented by:

41 / 50

41. A functional protein contains two identical catalytic chains and one structurally different regulatory chain. Its quaternary assembly contains:

42 / 50

42. The correctly interpreted hydrolysis sequence is:

Option Initial material Intermediate product More complete products
A Protein Nucleotides Pentose, phosphate and bases
B Polynucleotide Mononucleotides Pentose, phosphate and bases
C Polysaccharide Amino acids Nitrogenous bases
D Polynucleotide Monosaccharides only Amino acids and phosphate

43 / 50

43. AA linear polynucleotide has a free (mathrm{5'})-phosphate group at one end and a free (mathrm{3'})-hydroxyl group at the other. Each pentose sugar contains hydrogen rather than a hydroxyl group at (mathrm{C_{2'}}). The correct structural interpretation is:

44 / 50

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

45 / 50

45. Many enzymes are classified chemically as:

46 / 50

46. Consider the following statements about fructose.
Statement I: It is a monosaccharide.
Statement II: Its open-chain form contains a ketonic group at .
Statement III: It contains six carbon atoms and is therefore a hexose.
The acceptable statements are:

47 / 50

47. A maltose sample is completely hydrolysed to glucose. Using and , the mass of glucose formed is:

48 / 50

48. Adding an enzyme to a reversible reaction that is not at equilibrium will generally:

49 / 50

49. A polypeptide contains several -helices and -sheets arranged into a compact functional shape. The compact functional shape represents:

50 / 50

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

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