NUMS — Medical & Dental Entry Test

NUMS Biology — Biological Molecules and Enzymes (Set 5)

Take the NUMS Biology — Biological Molecules and Enzymes (Set 5): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review.…

What this test covers

Sample questions with answers

The first 5 questions of this test, with the correct answer and explanation.

  1. Q1. A single polypeptide chain is built from 120 amino acids joined end to end with no branching. How many water molecules are released while its primary structure is formed?

    • A. 121
    • B. 119 ✓
    • C. 120
    • D. 118

    Answer: B. 119

    Explanation: Each peptide bond releases one water. A linear chain of n residues has n − 1 bonds, so 120 − 1 = 119.

  2. Q2. A DNA fragment of 1,000 base pairs contains 280 adenine nucleotides. How many hydrogen bonds hold its two strands together (2 per A–T pair, 3 per G–C pair)?

    • A. 2,720 ✓
    • B. 2,560
    • C. 2,840
    • D. 2,280

    Answer: A. 2,720

    Explanation: A = T = 280 pairs; G = C = 720 pairs. Hydrogen bonds = 280×2 + 720×3 = 560 + 2160 = 2,720.

  3. Q3. An enzyme has Km = 3 mM and Vmax = 50 μmol/min. At what substrate concentration will its initial rate be 40 μmol/min?

    • A. 12 mM ✓
    • B. 9 mM
    • C. 6 mM
    • D. 15 mM

    Answer: A. 12 mM

    Explanation: v = Vmax[S]/(Km+[S]); v/Vmax = 40/50 = 0.8 = [S]/(3+[S]), so 0.8×3 = 0.2[S] and [S] = 12 mM (four times Km).

  4. Q4. An enzyme is assayed with and without an inhibitor. At 1 mM substrate the rate falls from 50 to 25 units with the inhibitor, but at 200 mM substrate both assays give almost identical rates near the maximum. Which inhibitor behaviour does this show?

    • A. Non-competitive inhibition, with Km raised permanently
    • B. Competitive inhibition, overcome by excess substrate ✓
    • C. Non-competitive inhibition, overcome by excess substrate
    • D. Competitive inhibition, with Vmax lowered permanently

    Answer: B. Competitive inhibition, overcome by excess substrate

    Explanation: If a very high substrate concentration restores the normal maximum rate, Vmax is unchanged and only apparent Km rises. This is competitive inhibition, in which substrate and inhibitor compete for the same active site.

  5. Q5. In a pathway A → B → C → D, the final product D normally slows enzyme 1 by binding to a site away from its active site. A mutant enzyme 1 can no longer bind D at that site but still catalyses A → B. What is the expected result?

    • A. D keeps being made, and may accumulate beyond need ✓
    • B. D production falls because the active site is blocked
    • C. D accumulates only if substrate A runs out first
    • D. D production stops because enzyme 1 is inactive

    Answer: A. D keeps being made, and may accumulate beyond need

    Explanation: Allosteric feedback inhibition is lost, so enzyme 1 is never switched off by D and the pathway runs on regardless of the level of D.

The other 25 questions

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  1. Q6. The rate of an enzyme-catalysed reaction at 25 °C is 4 units. Assuming a temperature coefficient (Q10) of 2 and no denaturation, what rate is expected at 45 °C?

    • A. 16 units
    • B. 32 units
    • C. 12 units
    • D. 8 units
  2. Q7. A starch solution gives a blue-black colour with iodine and no change with Benedict's reagent. After incubation with saliva, it gives no blue-black colour but a brick-red precipitate on heating with Benedict's reagent. What explains this?

    • A. Saliva oxidised starch and made it a reducing polysaccharide
    • B. Salivary amylase hydrolysed starch to non-reducing sucrose
    • C. Salivary lipase converted starch into reducing fatty acids
    • D. Salivary amylase hydrolysed starch to reducing sugar (maltose)
  3. Q8. A sugar solution gives no colour change with Benedict's reagent. After boiling with dilute HCl, neutralising, and re-testing, it gives a brick-red precipitate. The sugar was most probably

    • A. sucrose, hydrolysed to glucose and fructose
    • B. lactose, hydrolysed to glucose and galactose
    • C. maltose, hydrolysed to two glucose units
    • D. starch, hydrolysed to many glucose units
  4. Q9. A child with bleeding gums and slow wound healing has collagen that cannot be properly cross-linked because proline residues are not hydroxylated. Which dietary deficiency is the most likely cause?

    • A. Vitamin C
    • B. Vitamin K
    • C. Vitamin D
    • D. Vitamin A
  5. Q10. How many different tripeptides could be assembled if each of the three positions can be occupied by any one of the 20 common amino acids, repeats allowed?

    • A. 6,840
    • B. 1,140
    • C. 8,000
    • D. 400
  6. Q11. A cell makes 5 triglyceride molecules from glycerol and fatty acids. How many water molecules are released in total?

    • A. 5
    • B. 20
    • C. 15
    • D. 10
  7. Q12. Olive oil is liquid at room temperature while butter fat is solid. Which structural difference best accounts for this?

    • A. cis double bonds kink the chains, stopping close packing
    • B. shorter glycerol backbones in oil weaken ester bonds
    • C. oils contain more phosphate groups that repel each other
    • D. trans double bonds straighten the chains and loosen packing
  8. Q13. The waxy layer on a leaf surface is made of long-chain fatty acids joined to long-chain alcohols. Which bond type joins these two components?

    • A. Ester bond
    • B. Glycosidic bond
    • C. Phosphodiester bond
    • D. Peptide bond
  9. Q14. A person running on a hot day loses heat as sweat evaporates from the skin. Which property of water makes this cooling so effective?

    • A. High density at 4 °C, since cold water sinks from the skin
    • B. High latent heat of vaporisation, since many hydrogen bonds must break
    • C. High surface tension, since the film clings to skin pores
    • D. High specific heat, since it absorbs heat without changing state
  10. Q15. A rabbit survives on grass rich in cellulose although its own cells make no cellulase. Cellulose can be digested because

    • A. gut microbes make amylase that cuts its α-1,4 linkages
    • B. gut microbes make cellulase that cuts its β-1,4 linkages
    • C. the rabbit's maltase cuts its α-1,6 linkages slowly
    • D. the rabbit's amylase cuts its β-1,4 linkages slowly
  11. Q16. Glycogen can be mobilised faster than a long unbranched glucose polymer of the same size. The main reason is that glycogen has

    • A. many branch ends where enzymes release glucose together
    • B. many β-1,4 bonds that break more easily than α bonds
    • C. a hydrophobic core that keeps enzymes bound longer
    • D. fewer glucose units so each chain is quickly used
  12. Q17. The exoskeleton of an insect and the cell wall of a mushroom both contain the same structural polysaccharide. It is made of

    • A. galactose units only
    • B. β-glucose units only
    • C. amino acid units
    • D. N-acetylglucosamine units
  13. Q18. A nucleic acid sample analysed by chemists has the base composition A 23%, U 27%, G 31%, C 19%. What is the best description?

    • A. Single-stranded RNA, since U is present and A ≠ U
    • B. Single-stranded DNA, since U replaces T on some strands
    • C. Double-stranded DNA, since the bases pair in a ratio
    • D. Double-stranded RNA, since A and U are both present
  14. Q19. Sample X of DNA contains 30% G + C and sample Y contains 60% G + C. When both are heated slowly in solution, which statement is correct?

    • A. X melts at a higher temperature because A–T pairs have three hydrogen bonds
    • B. Y melts at a higher temperature because G–C pairs have three hydrogen bonds
    • C. both melt at the same temperature because length alone decides it
    • D. Y melts at a lower temperature because G–C pairs break more easily
  15. Q20. Hexokinase transfers a phosphate group from ATP to glucose. To which class of enzymes does it belong?

    • A. Hydrolase
    • B. Ligase
    • C. Transferase
    • D. Isomerase
  16. Q21. A person on a diet poor in niacin shows impaired activity of many dehydrogenases. Which coenzyme is directly affected?

    • A. Coenzyme A
    • B. NAD⁺
    • C. FAD
    • D. Pyridoxal phosphate
  17. Q22. The pancreas releases trypsinogen, not active trypsin. What is the biological advantage of this arrangement?

    • A. It lets trypsin work at the high pH of the stomach
    • B. It prevents the gland's own proteins being digested
    • C. It allows trypsin to bind a coenzyme in the duodenum
    • D. It increases the Vmax of trypsin once it is activated
  18. Q23. A mutation changes an amino acid far from an enzyme's active site, yet the enzyme loses activity. The most likely explanation is that the mutation

    • A. changed the peptide bond so substrate cannot bind
    • B. altered the folding so the active-site shape changed
    • C. added a cofactor that blocks the active site directly
    • D. raised the activation energy of every reaction in the cell
  19. Q24. An enzyme is heated at 70 °C for 10 minutes, then cooled to 37 °C. Its activity does not return. The best explanation is that

    • A. its peptide bonds were hydrolysed so it lost its amino acids
    • B. its substrate lost kinetic energy on cooling to 37 °C
    • C. its coenzyme was converted to an inhibitor at 70 °C
    • D. its tertiary structure and active site were irreversibly altered
  20. Q25. Reaction rate is measured at a fixed, limited substrate concentration while the enzyme concentration is raised step by step. The rate first rises in proportion and then stops rising. What is the best interpretation?

    • A. The active sites have been blocked by excess enzyme
    • B. Substrate has become limiting, so extra enzyme has nothing to act on
    • C. The temperature has risen above the optimum
    • D. The enzyme has become saturated with product
  21. Q26. Glycine has an isoelectric point of about 6. If it is placed in an electric field at pH 2, which statement is correct?

    • A. It carries a net positive charge and moves to the cathode
    • B. It carries no net charge and does not move
    • C. It carries a net negative charge and moves to the cathode
    • D. It carries a net positive charge and moves to the anode
  22. Q27. Silk fibroin is strong but hardly stretches, while hair keratin stretches under tension. Which structural feature explains the silk property?

    • A. Mostly disulfide-bonded globular domains with no hydrogen bonds
    • B. Mostly triple helices of glycine and hydroxyproline
    • C. Mostly extended β-pleated sheets stabilised between strands
    • D. Mostly tight α-helices stabilised along each strand
  23. Q28. A haemoglobin molecule has four polypeptide chains, each with one haem group binding one O₂. How many O₂ molecules can 250 fully saturated haemoglobin molecules carry?

    • A. 1,000
    • B. 500
    • C. 250
    • D. 750
  24. Q29. A drug that blocks cholesterol synthesis in the body would be expected to reduce production of which pair of hormones?

    • A. Adrenaline and thyroxine
    • B. Oxytocin and ADH
    • C. Insulin and glucagon
    • D. Testosterone and cortisol
  25. Q30. Malonate is added to isolated mitochondria that are oxidising succinate. Malonate closely resembles succinate and binds the active site of succinate dehydrogenase. Which result is expected?

    • A. Succinate accumulates, and the block cannot be eased at all
    • B. Succinate accumulates, and the block eases if succinate is raised
    • C. Fumarate accumulates, and the block cannot be eased at all
    • D. Fumarate accumulates, and the block eases if succinate is raised