NUMS — Medical & Dental Entry Test

NUMS Biology — Biological Molecules and Enzymes (Set 6)

Take the NUMS Biology — Biological Molecules and Enzymes (Set 6): 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 = 80 µmol/min and Km = 2 mM. Assuming Michaelis–Menten kinetics, what is its reaction velocity when [S] = 6 mM?
- A. 40 µmol/min
- B. 60 µmol/min ✓
- C. 53 µmol/min
- D. 48 µmol/min
Answer: B. 60 µmol/min
Explanation: v = Vmax[S]/(Km+[S]) = 80×6/(2+6) = 60 µmol/min.
Q2. A Lineweaver–Burk plot (1/v against 1/[S]) for an uninhibited enzyme crosses the 1/v axis at 0.02 min per µmol. What is Vmax?
- A. 100 µmol per min
- B. 25 µmol per min
- C. 0.02 µmol per min
- D. 50 µmol per min ✓
Answer: D. 50 µmol per min
Explanation: The y-intercept equals 1/Vmax, so Vmax = 1/0.02 = 50 µmol per min.
Q3. An enzyme-catalysed reaction has a temperature coefficient (Q10) of 2 and runs at 2 units per minute at 20 °C. Assuming no denaturation, what is the expected rate at 40 °C?
- A. 4 units per minute
- B. 16 units per minute
- C. 8 units per minute ✓
- D. 6 units per minute
Answer: C. 8 units per minute
Explanation: Two 10 °C steps each double the rate: 2 × 2 × 2 = 8 units per minute.
Q4. A double-stranded DNA fragment has 400 base pairs, and 30% of its nucleotides are adenine. Taking two hydrogen bonds per A–T pair and three per G–C pair, how many hydrogen bonds hold the fragment together?
- A. 1,040
- B. 880
- C. 960 ✓
- D. 1,000
Answer: C. 960
Explanation: 800 nucleotides: A = T = 240, G = C = 160. Bonds = 240×2 + 160×3 = 480 + 480 = 960.
Q5. A purified virus yields nucleic acid with the base composition A 25%, T 33%, G 19%, C 23%. What does this indicate?
- A. The genome is double-stranded DNA, because A + T is close to G + C
- B. The genome is RNA, because it contains thymine
- C. The genome is double-stranded DNA with a sequencing error
- D. The genome is single-stranded DNA, because A does not equal T ✓
Answer: D. The genome is single-stranded DNA, because A does not equal T
Explanation: Chargaff's rules (A = T, G = C) hold only for double-stranded DNA. Thymine shows it is DNA; unequal A and T show one strand.

The other 25 questions

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Q6. Two DNA samples of equal length are heated slowly and their melting temperatures (Tm) measured. Sample X is 60% G+C and sample Y is 40% G+C. What is the expected result?
- A. Y has the higher Tm, because A–T pairs have three hydrogen bonds
- B. Y has the higher Tm, because A–T pairs stack more tightly
- C. Both have the same Tm, because their length is equal
- D. X has the higher Tm, because G–C pairs have three hydrogen bonds
Q7. A protein contains three polypeptide chains with 250 amino acid residues in total. How many peptide bonds does it contain (ignore disulfide bridges)?
- A. 253
- B. 249
- C. 250
- D. 247
Q8. An amino acid has a pI of 6.0. During electrophoresis at pH 8.0 it will:
- A. carry a net negative charge and move towards the cathode
- B. carry a net positive charge and move towards the anode
- C. carry a net negative charge and move towards the anode
- D. carry a net positive charge and move towards the cathode
Q9. A student boils sucrose with dilute HCl for two minutes and immediately adds Benedict's reagent and heats. The mixture stays blue. What is the most likely reason?
- A. Sucrose is not hydrolysed by acid, so no reducing sugar formed
- B. The acid was not neutralised, so the alkaline reagent failed
- C. Boiling destroyed the glucose and fructose produced, so none was left to react
- D. Benedict's reagent reacts only with starch, not with sugars, so no change was expected
Q10. Two fats are compared. Fat P has an iodine number of 120 and fat Q an iodine number of 40. Which conclusion is correct?
- A. P has more C=C bonds and is likely liquid at room temperature
- B. Q has more C=C bonds and is likely liquid at room temperature
- C. P has more C=C bonds and is likely solid at room temperature
- D. Q has more C=C bonds and is likely solid at room temperature
Q11. Phospholipids with two fatty-acid tails (as in membranes) tend to form bilayers rather than micelles in water mainly because:
- A. the two bulky tails make a cylindrical molecule that cannot pack into a small sphere
- B. the tails form covalent bonds with each other across the layer, locking it flat
- C. the two tails make the molecule fully hydrophilic, so it spreads flat in a single sheet
- D. the phosphate heads repel each other strongly and so prevent any curvature of the layer
Q12. The graph of velocity against substrate concentration for an enzyme is S-shaped (sigmoidal) rather than hyperbolic. This suggests that the enzyme:
- A. has a single active site and obeys simple Michaelis–Menten kinetics
- B. is a ribozyme that does not need an active site
- C. has several subunits and binds substrate cooperatively
- D. is permanently saturated at all substrate concentrations
Q13. In a pathway P → Q → R → S, each step is catalysed by enzymes E1, E2 and E3 respectively. Adding S to the system reduces E1 activity, while adding Q or R has no effect. This is best described as:
- A. irreversible poisoning of E2 by heavy metals present in the extract
- B. competitive inhibition of E3 by a molecule resembling its substrate
- C. enzyme induction by the end-product, raising synthesis of E1
- D. end-product (feedback) inhibition acting on the first enzyme
Q14. Treating a globular enzyme with β-mercaptoethanol, a reducing agent, abolishes its activity. Which level of structure is most directly disrupted?
- A. Primary structure, through breaking of peptide bonds
- B. Tertiary structure, through breaking of disulfide bridges
- C. Quaternary structure only, through loss of ionic bonds
- D. Secondary structure, through loss of all the hydrogen bonds
Q15. In a lake, ice forms at the surface and fish survive below it. Which property of water is mainly responsible?
- A. Ice is denser than liquid water, so it sinks and warms the lake
- B. Water molecules are non-polar and do not form hydrogen bonds
- C. Ice is less dense than liquid water, so it floats and insulates
- D. Water has a very low specific heat, so it never freezes fully
Q16. In paper chromatography, a pigment spot moves 3.2 cm while the solvent front moves 8.0 cm from the origin. What is the Rf value of the pigment?
- A. 0.40
- B. 0.25
- C. 2.50
- D. 4.80
Q17. In methanol poisoning, patients are treated with ethanol. Ethanol works because it:
- A. acts as a coenzyme that speeds methanol breakdown
- B. binds to a separate allosteric site and activates alcohol dehydrogenase
- C. competes with methanol for the active site of alcohol dehydrogenase
- D. denatures the enzyme so that methanol is not converted
Q18. Enzyme X loses all activity after dialysis of the extract, and activity returns on adding the dialysate. Enzyme Y keeps its activity after dialysis. Which inference is most reasonable?
- A. X is denatured by dialysis, so adding the dialysate renatures it again
- B. Y needs a small cofactor, whereas X needs none and is stable on its own
- C. X needs a loosely bound small cofactor; Y's is tightly bound or absent
- D. X is a larger protein than Y, so it passes out through the dialysis membrane
Q19. A student measures starch breakdown by amylase at 37 °C by timing the loss of iodine colour. Which control best shows that the colour change is due to the enzyme's catalytic activity?
- A. A tube containing amylase and starch kept on ice
- B. A tube containing previously boiled amylase and starch
- C. A tube containing starch and a double volume of iodine
- D. A tube containing amylase without starch
Q20. In a catalase experiment, oxygen collected was 0, 12, 20, 25 and 27 mL at 0, 30, 60, 90 and 120 s. What is the initial rate of oxygen production?
- A. 0.40 mL per s
- B. 0.23 mL per s
- C. 0.27 mL per s
- D. 0.20 mL per s
Q21. Enzymes A and B have the same Vmax, but Km values of 0.5 mM and 5 mM respectively. At [S] = 0.5 mM, which statement is correct?
- A. A works at 50% of Vmax, and B at about 9%
- B. Both work at 50% of Vmax
- C. A works at 100% of Vmax, and B at 10%
- D. A works at about 9% of Vmax, and B at 50%
Q22. A child with poor wound healing, bleeding gums and weak blood vessels has a diet lacking fresh fruit. Which molecular defect explains the weak connective tissue?
- A. Failure to form peptide bonds in keratin
- B. Loss of α-helix in haemoglobin due to heme deficiency
- C. Missing glycine in every third position of collagen
- D. Under-hydroxylated proline in collagen
Q23. In a globular protein dissolved in water, amino acids with non-polar side chains (such as valine and leucine) are mostly found:
- A. in the peptide backbone as polar groups that bind water molecules
- B. only at the N-terminal end, where the chain begins to fold inwards
- C. on the surface, forming hydrogen bonds with the surrounding water
- D. in the interior, away from water, through hydrophobic interactions
Q24. Starch and cellulose are both polymers of glucose, yet humans digest only one of them. The key structural reason is that:
- A. starch has α-1,4 links, cellulose β-1,4 links; human enzymes cut only α links
- B. starch is branched through ester bonds that cellulose entirely lacks, so only starch is cut
- C. cellulose has peptide bonds joining its sugar units together in chains
- D. starch is made of fructose units, which humans can readily absorb and use
Q25. An enzyme-catalysed reaction is run at a fixed small amount of enzyme, with substrate concentration raised far above the Km. Doubling the substrate again has almost no effect on rate. The best explanation is that:
- A. the substrate has become a competitive inhibitor
- B. the temperature coefficient has fallen to zero
- C. enzyme active sites are almost all occupied
- D. the enzyme has been denatured by excess substrate
Q26. An inhibitor lowers the Vmax of an enzyme to half but leaves its Km unchanged, and adding more substrate does not reverse its effect. Which type of inhibition is this?
- A. Non-competitive inhibition
- B. Feedback activation
- C. Competitive inhibition
- D. Uncompetitive inhibition at the active site
Q27. A triglyceride is formed from one glycerol and three fatty acid molecules. How many water molecules are released when 100 triglyceride molecules are synthesised?
- A. 200
- B. 100
- C. 300
- D. 400
Q28. Casein, the main protein of milk, binds phosphate groups to serine side chains and so is classified as:
- A. a lipoprotein, which is a fibrous protein
- B. a glycoprotein, which is a simple protein
- C. a phosphoprotein, which is a conjugated protein
- D. a nucleoprotein, which is a simple globular protein
Q29. A mutation changes an enzyme's active-site residue so that it can no longer bind its substrate, yet its overall shape is unchanged. Compared with the normal enzyme, the mutant would show:
- A. a much higher Km or negligible activity, as substrate binding is lost
- B. a higher Vmax and the same Km, because binding is improved
- C. unchanged kinetics, as only the overall tertiary structure matters for catalysis
- D. a lower Km and the same Vmax, because the shape is intact
Q30. On a Lineweaver–Burk plot, a competitive inhibitor is added to an enzyme. Compared with the uninhibited line, the inhibited line shows:
- A. the same slope but a higher y-intercept
- B. the same x-intercept but a flatter slope
- C. the same y-intercept but a steeper slope
- D. a lower y-intercept and the same x-intercept