NUMS — Medical & Dental Entry Test

NUMS Biology — Biological Molecules and Enzymes (Set 4)

Take the NUMS Biology — Biological Molecules and Enzymes (Set 4): 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. An enzyme has Vmax = 120 μmol/min and Km = 2 mM. What is its initial rate at a substrate concentration of 6 mM (Michaelis–Menten kinetics)?
- A. 60 μmol/min
- B. 100 μmol/min
- C. 90 μmol/min ✓
- D. 80 μmol/min
Answer: C. 90 μmol/min
Explanation: v = Vmax[S]/(Km + [S]) = 120 × 6/(2 + 6) = 90 μmol/min.
Q2. On a Lineweaver–Burk plot, the lines for an enzyme with and without an inhibitor intersect on the y-axis, but the line with inhibitor has a steeper slope. The inhibitor is most likely
- A. competitive, because Vmax is unchanged while apparent Km increases ✓
- B. non-competitive, because Vmax falls while Km is unchanged
- C. competitive, because Km is unchanged while Vmax increases
- D. uncompetitive, because both Vmax and Km fall
Answer: A. competitive, because Vmax is unchanged while apparent Km increases
Explanation: The y-intercept is 1/Vmax; an unchanged intercept means Vmax is unchanged. A steeper slope (Km/Vmax) means a larger apparent Km, the signature of competitive inhibition.
Q3. A single polypeptide of 150 amino acids is held in shape by 2 disulfide bridges. How many peptide bonds are present in the chain?
- A. 149 ✓
- B. 148
- C. 152
- D. 150
Answer: A. 149
Explanation: A single unbranched chain of n residues has n − 1 peptide bonds = 149. Disulfide bridges are side-chain bonds, not peptide bonds.
Q4. How many different tripeptides can be built from the 20 standard amino acids if each amino acid may be used any number of times?
- A. 6840
- B. 400
- C. 1140
- D. 8000 ✓
Answer: D. 8000
Explanation: Each of the three positions can hold any of 20 amino acids: 20³ = 8000. (6840 = 20 × 19 × 18 would forbid repeats.)
Q5. A colourless solution gives no colour change with Benedict's reagent. After boiling with dilute HCl, neutralising with NaHCO₃ and retesting, a brick-red precipitate forms. The solution most likely contained
- A. maltose
- B. sucrose ✓
- C. glucose
- D. lactose
Answer: B. sucrose
Explanation: Sucrose is a non-reducing sugar (the glycosidic bond uses both anomeric carbons). Acid hydrolysis releases glucose and fructose, which are reducing sugars. Glucose, maltose and lactose would be positive without hydrolysis.

The other 25 questions

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Q6. Equal volumes of iodine solution are added to three tubes: tube 1 turns blue-black, tube 2 turns reddish-brown, tube 3 stays yellow-brown. Which set of contents is consistent?
- A. 1 = starch, 2 = cellulose, 3 = glycogen
- B. 1 = starch (amylose), 2 = glycogen, 3 = cellulose
- C. 1 = glycogen, 2 = starch, 3 = cellulose
- D. 1 = cellulose, 2 = starch, 3 = glycogen
Q7. An enzyme catalyses a reaction at 3 μmol/min at 20 °C. Assuming a temperature coefficient (Q₁₀) of 2 and no denaturation, what rate is expected at 30 °C?
- A. 4.5 μmol/min
- B. 9 μmol/min
- C. 12 μmol/min
- D. 6 μmol/min
Q8. A triglyceride is synthesised from glycerol and three fatty acids. How many water molecules are released and what type of bond forms?
- A. 3 water molecules; ester bonds
- B. 1 water molecule; ester bond
- C. 3 water molecules; glycosidic bonds
- D. 3 water molecules; peptide bonds
Q9. A DNA fragment is 1000 base pairs long and 20% of all its nucleotides are adenine. How many hydrogen bonds hold the two strands together?
- A. 2000
- B. 3200
- C. 2600
- D. 2400
Q10. In one strand of a DNA double helix, (A + G)/(T + C) = 0.8. What is this ratio in the complementary strand?
- A. 1.25
- B. 1.0
- C. 0.8
- D. 0.64
Q11. Which of the following is the best explanation for why humans cannot digest cellulose although they digest starch?
- A. They lack an enzyme able to hydrolyse β-1,4 glycosidic bonds
- B. Cellulose is made of fructose rather than glucose
- C. Starch is soluble in water whereas cellulose is not digestible by any acid
- D. Cellulose contains α-1,6 branches that block enzyme access
Q12. An inhibitor reduces the Vmax of an enzyme from 100 to 60 units but the Km stays the same. Adding extra substrate does not restore activity. The inhibitor acts by
- A. binding at a site other than the active site and lowering the amount of functional enzyme
- B. binding only to the enzyme–substrate complex and lowering Km
- C. denaturing the substrate
- D. competing with substrate for the active site
Q13. In a pathway P → Q → R → S catalysed by enzymes 1, 2 and 3, the end product S inhibits enzyme 1. If S accumulates in the cell, which change is expected?
- A. Levels of S rise further because of positive feedback
- B. Levels of Q rise because enzyme 2 is activated
- C. Levels of P rise and production of Q, R and S fall
- D. Levels of R rise because enzyme 3 is blocked
Q14. Trypsinogen is secreted by the pancreas and is converted to trypsin only in the duodenum. What is the main advantage of this arrangement?
- A. It supplies the cofactor trypsin needs to hydrolyse peptide bonds
- B. It raises the optimum pH of trypsin so that it can work in the duodenum
- C. It prevents the protease from digesting the pancreatic cells that make it
- D. It allows trypsin to act without a substrate bound at its active site
Q15. Which statement distinguishes a prosthetic group from a coenzyme?
- A. A prosthetic group is always an inorganic metal ion, whereas a coenzyme is always an organic vitamin-derived molecule
- B. A prosthetic group is a part of the substrate that is released, while a coenzyme is part of the enzyme protein chain
- C. A prosthetic group is tightly (often covalently) bound to the enzyme, whereas a coenzyme binds loosely and can leave
- D. A prosthetic group is used up and replaced in each reaction, whereas a coenzyme is recycled unchanged
Q16. A crude liver extract has 600 enzyme units in 20 mg of protein. After purification, the preparation has 600 units in 2 mg protein. What is the fold increase in specific activity?
- A. 5-fold
- B. 20-fold
- C. 100-fold
- D. 10-fold
Q17. An enzyme is already working at Vmax because the substrate is in excess. The amount of enzyme is now doubled. What happens to the initial reaction rate?
- A. It doubles only if temperature is also doubled
- B. It stays unchanged because substrate is limiting
- C. It roughly doubles, because more active sites are available
- D. It falls because enzymes compete for substrate
Q18. Pepsin (optimum pH about 2) is transferred to the duodenum where pH is about 8 and loses its activity. The best explanation is that
- A. the changed ionisation of R groups alters the shape of the enzyme, including its active site
- B. the substrate has already been hydrolysed in the stomach, so no protein is left for pepsin to act on
- C. pH 8 hydrolyses the peptide bonds of pepsin, so the enzyme is broken down into amino acids
- D. pepsin is converted back to inactive pepsinogen when the pH becomes alkaline in the duodenum
Q19. A water strider walks on a pond surface without sinking. Which property of water, and its basis, explains this?
- A. High polarity that makes water a poor solvent
- B. High specific heat from adhesion to the insect's legs
- C. Low density of liquid water caused by ionic bonds
- D. High surface tension from cohesion between molecules linked by hydrogen bonds
Q20. An α-helical segment of a protein has 36 amino acid residues with 3.6 residues per turn. How many complete turns does it contain?
- A. 9
- B. 36
- C. 10
- D. 12
Q21. In sickle-cell haemoglobin, glutamic acid at position 6 of the β-chain is replaced by valine. Which level of protein structure is directly altered by the mutation, and what is the consequence?
- A. Quaternary structure only; the α-chains are replaced
- B. Tertiary structure only; the amino acid sequence is unchanged
- C. Secondary structure only; the haem group is lost
- D. Primary structure; the haemoglobin molecules aggregate at low oxygen
Q22. Haemoglobin A is a tetramer with one haem per polypeptide chain. How many oxygen molecules can one haemoglobin molecule carry when fully saturated?
- A. 8
- B. 4
- C. 2
- D. 1
Q23. A few drops of ethanol extract of a food sample are added to water and the mixture turns milky-white. This result indicates
- A. lipid, because the dissolved lipid forms an emulsion in water
- B. starch, because the solution scatters light
- C. protein, because peptide bonds are present
- D. reducing sugar, because a precipitate forms
Q24. Phospholipids are shaken in water. Which arrangement is expected and why?
- A. A monolayer with hydrophobic heads facing the water and hydrophilic tails pointing inwards
- B. A bilayer or micelle with hydrophilic heads facing water and hydrophobic tails facing inward
- C. A random dispersion of single molecules, because phospholipids are fully soluble in water
- D. A layer of fatty-acid tails facing the water on both sides, with the heads hidden in the middle
Q25. Which of the following is NOT synthesised from cholesterol?
- A. Thyroxine
- B. Bile salts
- C. Oestrogen
- D. Vitamin D
Q26. A nucleotide contains ribose, uracil and three phosphate groups. It is best identified as
- A. UDP
- B. dUTP
- C. UTP
- D. dTTP
Q27. An uncatalysed reaction has an activation energy of 80 kJ/mol and ΔG = −20 kJ/mol. An enzyme lowers the activation energy to 50 kJ/mol. What is ΔG for the catalysed reaction?
- A. −30 kJ/mol
- B. +30 kJ/mol
- C. −20 kJ/mol
- D. −50 kJ/mol
Q28. An enzyme binds a stable analogue of the transition state far more tightly than it binds its substrate. This observation best supports the view that
- A. the active site is a rigid lock fitting only the substrate
- B. the substrate is covalently destroyed on binding
- C. the active site is complementary to the transition state, stabilising it
- D. enzymes change the equilibrium constant
Q29. Enzyme X has Km = 0.5 mM and enzyme Y has Km = 5 mM for the same substrate, with equal Vmax. Which conclusion is correct?
- A. X reaches half of its Vmax at a lower substrate concentration, so it has the higher affinity
- B. Y has the higher affinity because it needs a larger substrate concentration to reach half of its Vmax
- C. Both have the same affinity because Vmax is equal, since affinity is set only by Vmax
- D. X has the lower Vmax because its smaller Km means it saturates earlier and reaches a lower maximum rate
Q30. An enzyme converts glucose 6-phosphate to fructose 6-phosphate without adding or removing any atoms. It belongs to which class?
- A. Isomerases
- B. Ligases
- C. Hydrolases
- D. Transferases