NUMS — Medical & Dental Entry Test

NUMS Biology — Biological Molecules and Enzymes (Set 2)

Take the NUMS Biology — Biological Molecules and Enzymes (Set 2): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review.…

What this test covers

30 NUMS Biology

Sample questions with answers

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

Q1. A polysaccharide sample gives a blue-black colour with iodine and, on complete hydrolysis, yields only glucose. Microscopy shows unbranched helical chains. The sample is most likely:
- A. Chitin
- B. Cellulose
- C. Amylose ✓
- D. Amylopectin
Answer: C. Amylose
Explanation: Amylose is an unbranched α(1→4) glucose polymer that coils into a helix and traps iodine to give a blue-black colour; amylopectin is branched, cellulose does not form a helix, and chitin contains N-acetylglucosamine.
Q2. A linear polymer contains 500 glucose residues joined by glycosidic bonds. How many water molecules are released when it is synthesised from free glucose?
- A. 250
- B. 499 ✓
- C. 501
- D. 500
Answer: B. 499
Explanation: Each glycosidic bond formed by condensation releases one water molecule; a linear chain of n monomers has n − 1 bonds, so 500 − 1 = 499.
Q3. Sucrose does not give a positive Benedict's test, whereas maltose does. The best explanation is that in sucrose:
- A. Glucose is present in the β form only
- B. The glycosidic bond is a peptide bond
- C. There are three monosaccharide units
- D. The reducing groups of both glucose and fructose are used in the glycosidic bond ✓
Answer: D. The reducing groups of both glucose and fructose are used in the glycosidic bond
Explanation: Sucrose's α1→β2 linkage involves the anomeric carbons of both glucose and fructose, so no free reducing group remains; maltose keeps one free anomeric carbon.
Q4. Insects and fungi both contain a structural polysaccharide that resembles cellulose but has a nitrogen-containing group on each monomer. This polysaccharide is:
- A. Chitin ✓
- B. Glycogen
- C. Pectin
- D. Hemicellulose
Answer: A. Chitin
Explanation: Chitin is a β(1→4) polymer of N-acetylglucosamine, found in arthropod exoskeletons and fungal cell walls.
Q5. Saponification of a fat with sodium hydroxide produces soap and which other product?
- A. Glucose
- B. Cholesterol
- C. Amino acids
- D. Glycerol ✓
Answer: D. Glycerol
Explanation: Alkaline hydrolysis breaks the ester bonds of a triglyceride, giving sodium salts of fatty acids (soap) and glycerol.

The other 25 questions

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Q6. Per gram, fats release about twice as much energy as carbohydrates on oxidation. The main reason is that fats:
- A. Contain more phosphate groups
- B. Are oxidised anaerobically
- C. Contain a higher proportion of C–H bonds and less oxygen
- D. Are stored in larger cells
Q7. Which molecule is a steroid that serves as the precursor of sex hormones and vitamin D in humans?
- A. Prostaglandin
- B. Cholesterol
- C. Lecithin
- D. Triolein
Q8. Waxes on the leaves of desert plants mainly help by:
- A. Storing energy for winter
- B. Absorbing light for photosynthesis
- C. Speeding up gas exchange
- D. Reducing water loss through the cuticle
Q9. A tripeptide is formed from three different amino acids. How many peptide bonds and how many free terminal amino groups does it have?
- A. 3 peptide bonds and 0 free terminal amino groups
- B. 2 peptide bonds and 1 free terminal amino group
- C. 2 peptide bonds and 3 free terminal amino groups
- D. 3 peptide bonds and 1 free terminal amino group
Q10. The number of different tripeptides that can be made using the 20 standard amino acids (repetition allowed) is:
- A. 1140
- B. 400
- C. 60
- D. 8000
Q11. Keratin in hair and collagen in tendons are fibrous proteins. Compared with globular proteins like enzymes, fibrous proteins are generally:
- A. Soluble and catalytic
- B. Lacking secondary structure
- C. Made of nucleotides
- D. Insoluble in water and mainly structural
Q12. In a globular protein dissolved in water, amino acids with non-polar R groups such as leucine and valine are usually found:
- A. Spread evenly on the surface
- B. Only at the N-terminal end
- C. Bonded to water by hydrogen bonds
- D. Buried in the interior, away from water
Q13. Which bond can be broken by a reducing agent such as mercaptoethanol and is formed between two cysteine residues?
- A. Peptide bond
- B. Glycosidic bond
- C. Disulphide bridge
- D. Phosphodiester bond
Q14. A nucleotide consists of:
- A. A pentose sugar and a nitrogenous base only
- B. A pentose sugar, a nitrogenous base and a phosphate group
- C. An amino acid, a sugar and a phosphate group
- D. A hexose sugar, a base and a phosphate group
Q15. A double-stranded DNA fragment has 200 base pairs, of which 60 are A–T pairs. How many hydrogen bonds hold the two strands together?
- A. 600
- B. 400
- C. 460
- D. 540
Q16. A DNA sample with a high proportion of G–C pairs needs a higher temperature to separate its strands because G–C pairs:
- A. Are joined by covalent bonds
- B. Contain deoxyribose instead of ribose
- C. Are joined by three hydrogen bonds instead of two
- D. Are always found at the ends of the molecule
Q17. Which statement about the two strands of a DNA double helix is correct?
- A. Purines pair with purines to keep a constant width
- B. They are joined by phosphodiester bonds between bases
- C. They run antiparallel, one 5′→3′ and the other 3′→5′
- D. They are identical in base sequence
Q18. Sweating cools the body effectively because water has a high:
- A. Surface tension
- B. Latent heat of vaporisation
- C. Viscosity
- D. Density
Q19. A pond freezes on its surface in winter while fish survive below. This is possible because:
- A. Water has a low specific heat capacity
- B. Ice is less dense than liquid water and insulates the water beneath
- C. Fish produce heat by photosynthesis
- D. Ice is denser than water and sinks
Q20. An enzyme lowers the activation energy of a reaction. Which quantity does it NOT change?
- A. The overall free-energy change (ΔG) of the reaction
- B. The rate of the reaction
- C. The time taken to reach equilibrium
- D. The energy of the transition state
Q21. Enzyme X has a Km of 2 mM and enzyme Y has a Km of 10 mM for the same substrate. Which conclusion is correct?
- A. Both enzymes have equal affinity
- B. Enzyme X always has a higher Vmax
- C. Enzyme X has a higher affinity for the substrate
- D. Enzyme Y has a higher affinity for the substrate
Q22. Methanol poisoning is treated by giving ethanol, which competes for the active site of alcohol dehydrogenase. The ethanol works because:
- A. It binds an allosteric site and raises enzyme activity
- B. It reacts directly with methanol to form water
- C. It destroys the enzyme permanently
- D. It occupies the active site, so less methanol is converted to toxic formaldehyde
Q23. On a graph of rate against substrate concentration, a competitive inhibitor causes:
- A. A lower Vmax but the same Km
- B. A lower Vmax and lower Km
- C. A higher Vmax and lower Km
- D. The same Vmax but a higher apparent Km
Q24. Salivary amylase is active in the mouth but stops working in the stomach. The most likely reason is:
- A. The low pH of the stomach denatures it
- B. There is no starch in the stomach
- C. Pepsin activates amylase into a protease
- D. The stomach is too cold
Q25. An enzyme solution is kept at 0 °C for an hour and then warmed to 37 °C; it works normally. A second sample is kept at 80 °C and then cooled to 37 °C; it does not work. This shows that:
- A. The enzyme's optimum temperature is 0 °C
- B. Both low and high temperatures denature enzymes permanently
- C. High temperature only lowers kinetic energy
- D. Low temperature inactivates enzymes reversibly, while high temperature denatures them irreversibly
Q26. An enzyme converts glucose-6-phosphate into fructose-6-phosphate by rearranging atoms within the molecule. It belongs to the class:
- A. Transferases
- B. Ligases
- C. Hydrolases
- D. Isomerases
Q27. The non-protein part of an enzyme that is tightly and permanently bound to it, such as the haem group of catalase, is called a:
- A. Substrate
- B. Coenzyme
- C. Apoenzyme
- D. Prosthetic group
Q28. Which property of enzymes allows a very small amount to process a large amount of substrate?
- A. They are made of lipids
- B. They raise the activation energy
- C. They are not used up in the reaction
- D. They are destroyed after one reaction
Q29. An allosteric activator increases enzyme activity by:
- A. Breaking the enzyme into smaller subunits
- B. Binding to a site other than the active site and changing the active site's shape
- C. Supplying extra substrate to the enzyme
- D. Binding to the active site in place of the substrate
Q30. A food sample turns violet when sodium hydroxide and a few drops of dilute copper sulphate are added. The sample contains:
- A. Reducing sugar
- B. Lipid
- C. Protein
- D. Starch