NUMS — Medical & Dental Entry Test

NUMS Biology — Biological Molecules and Enzymes (Set 1)

Take the NUMS Biology — Biological Molecules and Enzymes (Set 1): 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 student heats a sample of egg albumin to 90 °C. Which level of protein structure is LEAST likely to be disrupted by this treatment?
- A. Tertiary structure
- B. Secondary structure
- C. Primary structure ✓
- D. Quaternary structure
Answer: C. Primary structure
Explanation: Heat breaks hydrogen, ionic and hydrophobic interactions but not the covalent peptide bonds that hold the amino-acid sequence (primary structure) together.
Q2. A disaccharide on hydrolysis yields two molecules of glucose joined originally by an α(1→4) glycosidic bond. The disaccharide is:
- A. Maltose ✓
- B. Lactose
- C. Cellobiose
- D. Sucrose
Answer: A. Maltose
Explanation: Maltose = glucose + glucose linked α(1→4). Cellobiose is the β(1→4) dimer; sucrose and lactose contain fructose and galactose respectively.
Q3. Why can humans digest starch but not cellulose, even though both are polymers of glucose?
- A. Starch is soluble in water while cellulose contains no glycosidic bonds
- B. Cellulose molecules are too branched for enzymes to reach
- C. Cellulose is made of fructose units, not glucose
- D. Human enzymes can hydrolyse α-glycosidic bonds but not β(1→4) bonds of cellulose ✓
Answer: D. Human enzymes can hydrolyse α-glycosidic bonds but not β(1→4) bonds of cellulose
Explanation: Both are glucose polymers; starch has α-linkages which amylase hydrolyses, whereas cellulose has β(1→4) linkages for which humans lack cellulase.
Q4. A triglyceride is formed from one glycerol and three fatty acids. How many molecules of water are released during its formation?
- A. Three ✓
- B. One
- C. Two
- D. Four
Answer: A. Three
Explanation: Each ester bond formed between a glycerol –OH and a fatty acid –COOH releases one water molecule; three ester bonds release three waters.
Q5. Which feature best explains why phospholipids spontaneously form bilayers in water?
- A. They contain three fatty-acid chains attached to glycerol
- B. They are entirely non-polar molecules
- C. They are linked by peptide bonds into long chains
- D. They have a hydrophilic phosphate head and two hydrophobic fatty-acid tails ✓
Answer: D. They have a hydrophilic phosphate head and two hydrophobic fatty-acid tails
Explanation: Amphipathic phospholipids orient heads toward water and tails away from it, which naturally produces a bilayer.

The other 25 questions

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Q6. In a DNA sample, 18% of the bases are adenine. What percentage of the bases are cytosine?
- A. 64%
- B. 18%
- C. 32%
- D. 36%
Q7. Which statement correctly distinguishes RNA from DNA?
- A. RNA is always double-stranded with complementary base pairing
- B. RNA lacks phosphate groups in its backbone
- C. RNA contains deoxyribose and thymine
- D. RNA contains ribose and uracil, and is usually single-stranded
Q8. Adding more substrate to an enzyme reaction restores the original rate of reaction in the presence of an inhibitor. The inhibitor is most likely:
- A. An allosteric activator
- B. Irreversible
- C. Non-competitive
- D. Competitive
Q9. A non-competitive inhibitor is added to an enzyme-catalysed reaction. Which effect on the kinetics is expected?
- A. Vmax increases while Km decreases
- B. Vmax decreases while Km remains unchanged
- C. Vmax unchanged while Km increases
- D. Both Vmax and Km increase
Q10. The rate of an enzyme reaction stops rising as substrate concentration increases further. The best explanation is that:
- A. All active sites are occupied, so enzyme concentration has become limiting
- B. The substrate has denatured the enzyme
- C. The activation energy has risen
- D. The products are destroying the substrate
Q11. An enzyme from a hot-spring bacterium has an optimum temperature of 75 °C. Compared with human salivary amylase, it most likely has:
- A. A lower activation energy requirement for denaturation
- B. No active site
- C. More stabilising bonds such as disulphide and ionic bonds in its tertiary structure
- D. Only primary structure
Q12. Pepsin works best at pH 2 while trypsin works best at about pH 8. If pepsin enters the duodenum, its activity falls mainly because:
- A. Temperature in the duodenum is too low
- B. Pepsin needs a coenzyme found only in the stomach
- C. The change in pH alters ionic charges in the active site, changing its shape
- D. Pepsin is digested instantly by bile
Q13. Which pairing of cofactor type and example is correct?
- A. Coenzyme — Cl⁻ ion
- B. Activator — FAD permanently bound
- C. Coenzyme — NAD⁺
- D. Prosthetic group — Mg²⁺ loosely attached
Q14. The induced-fit model differs from the lock-and-key model because it proposes that:
- A. Enzymes can bind any substrate
- B. The enzyme is used up during the reaction
- C. The active site is rigid and exactly complementary before binding
- D. The active site changes shape slightly on substrate binding to fit it more closely
Q15. Enzymes speed up reactions by:
- A. Lowering the activation energy needed for the reaction
- B. Increasing the free-energy difference between reactants and products
- C. Shifting the equilibrium towards the products
- D. Raising the temperature of the reaction mixture
Q16. In the end-product (feedback) inhibition of a metabolic pathway, the final product usually:
- A. Binds to an allosteric site of the first enzyme and inhibits it
- B. Activates all enzymes of the pathway
- C. Denatures the substrate of the first enzyme
- D. Binds to the active site of the last enzyme irreversibly
Q17. A heavy-metal ion like Hg²⁺ permanently inactivates an enzyme by binding to its –SH groups. This is an example of:
- A. Irreversible inhibition
- B. Competitive reversible inhibition
- C. Allosteric activation
- D. Feedback inhibition
Q18. Which bond links the monomers in each macromolecule correctly?
- A. Fat — peptide bond; starch — phosphodiester bond
- B. Protein — ester bond; DNA — glycosidic bond
- C. DNA — peptide bond; glycogen — ester bond
- D. Protein — peptide bond; polysaccharide — glycosidic bond
Q19. Glycogen is a better energy store for animals than amylose mainly because glycogen:
- A. Contains β-glycosidic bonds
- B. Has a higher energy content per gram than fat
- C. Is soluble and therefore osmotically active
- D. Is highly branched, giving many ends for rapid release of glucose
Q20. A fatty acid with one C=C double bond in its hydrocarbon chain is described as:
- A. Monounsaturated
- B. Hydrogenated
- C. Polyunsaturated
- D. Saturated
Q21. Unsaturated fats are usually liquid at room temperature because:
- A. They have longer chains than saturated fats
- B. They form more hydrogen bonds between molecules
- C. Cis double bonds create kinks that prevent tight packing of the chains
- D. They contain no glycerol
Q22. During electrophoresis at a pH equal to its isoelectric point, an amino acid does not migrate towards either electrode. This is because it:
- A. Has lost its carboxyl group
- B. Carries a net negative charge
- C. Exists as a zwitterion with zero net charge
- D. Carries a net positive charge
Q23. Which interaction stabilises the α-helix of a protein?
- A. Disulphide bridges between adjacent residues
- B. Ionic bonds between phosphate groups
- C. Peptide bonds between R-groups
- D. Hydrogen bonds between C=O and N–H groups of the backbone
Q24. Haemoglobin consists of four polypeptide chains. This level of organisation is called:
- A. Tertiary structure
- B. Primary structure
- C. Quaternary structure
- D. Secondary structure
Q25. A single amino acid substitution (glutamic acid → valine) in the β-chain of haemoglobin causes sickle-cell disease. This shows that:
- A. Amino acid sequence has no effect on folding
- B. Only quaternary structure determines function
- C. A change in primary structure can alter the final shape and function of a protein
- D. Secondary structure is independent of R-groups
Q26. Which reagent would detect the presence of reducing sugars such as glucose in a food sample?
- A. Biuret reagent, giving a violet colour
- B. Iodine solution, giving a blue-black colour
- C. Sudan III, giving a red-stained layer
- D. Benedict's solution, giving a brick-red precipitate on heating
Q27. Water is an excellent coolant in living organisms primarily because of its:
- A. Non-polar nature
- B. Low density as a liquid
- C. High latent heat of vaporisation
- D. Low specific heat capacity
Q28. Which of the following is a ketose sugar?
- A. Galactose
- B. Fructose
- C. Glucose
- D. Ribose
Q29. An enzyme that removes a phosphate group from a substrate by adding water across the bond belongs to which class?
- A. Isomerases
- B. Ligases
- C. Oxidoreductases
- D. Hydrolases
Q30. If an enzyme's coenzyme is derived from a vitamin, a dietary deficiency of that vitamin would most directly cause:
- A. Denaturation of all proteins in the cell
- B. Increased synthesis of the apoenzyme into a more active form
- C. Reduced activity of the enzyme because the holoenzyme cannot form
- D. Change in the enzyme's primary structure