NUMS — Medical & Dental Entry Test

NUMS Biology — Bioenergetics (Set 7)

Take the NUMS Biology — Bioenergetics (Set 7): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review. Covers 30 NUMS…

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. Take one glucose molecule through complete aerobic respiration. How many ATP molecules are made by substrate-level phosphorylation alone (net, in glycolysis and the Krebs cycle together, counting GTP as ATP)?
- A. 8
- B. 4 ✓
- C. 2
- D. 6
Answer: B. 4
Explanation: Glycolysis gives a net of 2 ATP and the Krebs cycle gives 1 ATP per turn, so 2 turns give 2 more: 2 + 2 = 4.
Q2. Assume each NADH gives 3 ATP and each FADH₂ gives 2 ATP at the electron transport chain, with no shuttle costs. The oxidation of one glucose produces 10 NADH and 2 FADH₂. How many ATP come from oxidative phosphorylation?
- A. 34 ✓
- B. 32
- C. 30
- D. 38
Answer: A. 34
Explanation: 10 × 3 + 2 × 2 = 34 ATP by oxidative phosphorylation, under the stated convention.
Q3. Isolated mitochondria in a buffer take up no oxygen when glucose is added, but take up oxygen quickly when pyruvate is added. What does this show?
- A. The Krebs cycle needs glucose to be converted in the matrix first
- B. Glycolysis occurs outside mitochondria, which oxidise pyruvate ✓
- C. The electron transport chain cannot use oxygen from a buffer
- D. Glycolysis occurs inside mitochondria, which lack a pyruvate enzyme
Answer: B. Glycolysis occurs outside mitochondria, which oxidise pyruvate
Explanation: Glycolysis enzymes are in the cytosol. Mitochondria take in pyruvate for the link reaction and the Krebs cycle, which then feed the electron transport chain.
Q4. Malonate competitively inhibits succinate dehydrogenase. In a mitochondrial suspension oxidising pyruvate, which compound builds up?
- A. Fumarate
- B. Succinate ✓
- C. Citrate
- D. Oxaloacetate
Answer: B. Succinate
Explanation: Succinate dehydrogenase converts succinate to fumarate. Blocking it makes the substrate, succinate, accumulate and the products downstream fall.
Q5. In a person with severe thiamine (vitamin B₁) deficiency, pyruvate and lactate accumulate in the blood. Which step is most directly affected?
- A. The Calvin cycle, because RuBisCO needs thiamine pyrophosphate
- B. The electron transport chain, because cytochrome oxidase needs thiamine
- C. The link reaction, because pyruvate dehydrogenase needs thiamine pyrophosphate ✓
- D. Glycolysis, because hexokinase needs thiamine pyrophosphate
Answer: C. The link reaction, because pyruvate dehydrogenase needs thiamine pyrophosphate
Explanation: Thiamine pyrophosphate is a coenzyme of pyruvate dehydrogenase. Without it pyruvate cannot enter the Krebs cycle and is diverted to lactate.

The other 25 questions

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Q6. A chemical makes the inner mitochondrial membrane leaky to protons. Cells given this chemical consume oxygen faster but make much less ATP and give off heat. Why?
- A. Electron transport stops, so oxygen is no longer used and ATP falls
- B. Glycolysis stops completely, so the cell must use more oxygen
- C. ATP synthase is destroyed, so electrons cannot reach the oxygen
- D. The proton gradient collapses and energy escapes as heat, not ATP
Q7. Cyanide blocks cytochrome oxidase, the last carrier in the electron transport chain. Shortly afterwards, how will the matrix NADH level and Krebs cycle activity change?
- A. NADH rises and Krebs cycle speeds up, because more NAD⁺ is made
- B. NADH rises and Krebs cycle slows, because NAD⁺ is not regenerated
- C. NADH falls and Krebs cycle slows, because less glucose is present
- D. NADH falls and Krebs cycle speeds up, because oxygen is in excess
Q8. A leaf is supplied with water containing the heavy isotope ¹⁸O (H₂¹⁸O) and ordinary CO₂ in bright light. Almost all of the labelled oxygen atoms would be released as part of:
- A. the glucose built in the Calvin cycle
- B. the carbon dioxide given off by the leaf
- C. the oxygen gas released by the leaf
- D. the NADPH formed at photosystem I
Q9. Palmitic acid (C₁₆H₃₂O₂) is oxidised completely in a respirometer: C₁₆H₃₂O₂ + 23O₂ → 16CO₂ + 16H₂O. What is the RQ?
- A. 0.70
- B. 1.44
- C. 1.00
- D. 0.87
Q10. A culture of germinating seeds uses 10 mL of oxygen and gives off 12 mL of carbon dioxide in an hour. What is the RQ, and what does it suggest?
- A. 0.83, which suggests protein is the only substrate
- B. 0.7, which suggests mostly fat is being oxidised
- C. 1.2, which suggests some anaerobic respiration is occurring
- D. 1.0, which suggests only glucose is oxidised aerobically
Q11. A capillary respirometer has a bore of radius 0.5 mm. The liquid index moves 12 mm in one hour as germinating seeds take up oxygen (CO₂ absorbed by KOH). Using π = 3.14, what volume of oxygen was used?
- A. 37.7 mm³
- B. 18.8 mm³
- C. 4.71 mm³
- D. 9.42 mm³
Q12. Yeast ferments 180 g of glucose (1 mol) completely to ethanol and carbon dioxide. What volume of carbon dioxide is released at STP (22.4 L per mole)?
- A. 22.4 L
- B. 44.8 L
- C. 134.4 L
- D. 67.2 L
Q13. During intense exercise a muscle converts pyruvate to lactate. What is the key biochemical benefit of this step for the cell?
- A. It generates extra ATP from the lactate formed
- B. It removes carbon dioxide that has built up in muscle
- C. It converts NADH into FADH₂ for the electron transport chain
- D. It regenerates NAD⁺ for continued glycolysis and ATP formation
Q14. A fatty acid with 18 carbon atoms is broken down by β-oxidation, which removes two carbons as acetyl-CoA in each cycle. How many acetyl-CoA molecules and how many cycles of β-oxidation result?
- A. 8 acetyl-CoA and 8 cycles
- B. 9 acetyl-CoA and 9 cycles
- C. 18 acetyl-CoA and 9 cycles
- D. 9 acetyl-CoA and 8 cycles
Q15. Heavy water with ¹⁸O is supplied to a photosynthesising alga in one experiment, and C¹⁸O₂ in another. In which product does ¹⁸O appear in each case?
- A. From H₂¹⁸O in the carbohydrate, and from C¹⁸O₂ in the released oxygen gas
- B. From both sources in the released oxygen gas only
- C. H₂¹⁸O labels the released oxygen; C¹⁸O₂ labels water and carbohydrate
- D. From both sources in the carbohydrate only
Q16. A mutant plant lacks photosystem II but has a working photosystem I and electron carriers. Which process can still occur?
- A. Cyclic photophosphorylation, giving ATP without NADPH or oxygen
- B. Complete Calvin cycle, giving glucose from CO₂
- C. Non-cyclic photophosphorylation, giving ATP, NADPH and oxygen
- D. Photolysis of water, giving oxygen and protons
Q17. Which reaction centre gets its electrons replaced directly from the splitting of water?
- A. P680 of photosystem I
- B. P700 of photosystem II
- C. P680 of photosystem II
- D. P700 of photosystem I
Q18. A leaf kept in the dark at room temperature takes up 2 µL of oxygen per hour. At 400 lux it gives out 6 µL of oxygen per hour (net). What is the gross rate of oxygen production by photosynthesis at 400 lux?
- A. 12 µL per hour
- B. 6 µL per hour
- C. 8 µL per hour
- D. 4 µL per hour
Q19. Photosynthesis in a pond plant is measured at two CO₂ levels. At 0.03% CO₂ the rate stops rising above 2000 lux at a value of 6; at 0.10% CO₂ it plateaus at 14. What limits the rate at the plateau for the 0.03% CO₂ plant?
- A. Water potential
- B. Carbon dioxide level
- C. Chlorophyll amount
- D. Light intensity
Q20. A lamp is moved from 10 cm to 20 cm from a pond plant that produces 40 bubbles per minute at 10 cm. Light intensity varies as 1/distance², and rate is proportional to intensity when light is limiting. What rate is expected at 20 cm?
- A. 10 bubbles per minute
- B. 20 bubbles per minute
- C. 30 bubbles per minute
- D. 5 bubbles per minute
Q21. In a chromatogram, a pigment spot has moved 4.5 cm while the solvent front moved 9.0 cm. Reference Rf values: carotene 0.95, chlorophyll a 0.55, chlorophyll b 0.50, xanthophyll 0.35. Which pigment is it, and what is its Rf?
- A. Chlorophyll b, with Rf 0.50
- B. Carotene, with Rf 0.95
- C. Chlorophyll a, with Rf 0.55
- D. Xanthophyll, with Rf 0.35
Q22. To make two molecules of glucose through the Calvin cycle, the cycle needs 18 ATP and 12 NADPH per glucose. How many ATP and NADPH are used in total?
- A. 18 ATP and 24 NADPH
- B. 18 ATP and 12 NADPH
- C. 36 ATP and 12 NADPH
- D. 36 ATP and 24 NADPH
Q23. On a hot, dry day a C₃ plant closes its stomata. Photorespiration increases in its leaves. Which change inside the leaf is the main reason?
- A. The CO₂ to O₂ ratio rises, so RuBisCO binds more carbon dioxide
- B. Light intensity falls, so ATP formation stops
- C. The O₂ to CO₂ ratio rises, so RuBisCO binds more oxygen
- D. Water enters the chloroplast, so stroma pH falls
Q24. A CAM plant leaf is titrated with alkali for acidity. When is the titratable acidity of the cell sap highest, and why?
- A. At midnight, because malic acid is converted to glucose then
- B. At noon, because stomata are open and CO₂ enters
- C. At dusk, because malic acid is produced by the light reactions
- D. At dawn, because malic acid stored overnight has not yet been used
Q25. Which statement about the action spectrum of photosynthesis is correct?
- A. It is identical to the absorption spectrum of chlorophyll a alone
- B. It has a single peak in far-red light, where oxygen is made
- C. It shows the highest rate in green light, where absorption is greatest
- D. It follows the combined pigment absorption, with peaks in blue and red
Q26. In the inner mitochondrial membrane, ATP synthase consists of F₀ and F₁ parts. Which statement is correct?
- A. F₀ is the catalytic head in the matrix and F₁ is a channel for oxygen
- B. F₀ pumps protons into the matrix and F₁ breaks down ATP
- C. F₀ is in the intermembrane space and F₁ carries electrons
- D. F₀ is a membrane proton channel and F₁ is the catalytic head
Q27. A destarched plant has one leaf partly covered with black paper strips and is placed in bright light for a day. The leaf is decolourised in hot ethanol and tested with iodine. What will be seen?
- A. Blue-black where light reached the leaf, brown under the strips
- B. Blue-black everywhere, because starch moves across the leaf
- C. Blue-black under the strips, brown where light reached the leaf
- D. Brown everywhere, because iodine cannot enter the leaf
Q28. How many CO₂ molecules are released when three glucose molecules are oxidised completely through the link reaction and Krebs cycle? (Ignore any CO₂ from fermentation.)
- A. 12
- B. 18
- C. 9
- D. 6
Q29. The hydrolysis of ATP to ADP and phosphate releases 30.6 kJ per mole under standard conditions. How much energy is released when 0.2 mol of ATP is hydrolysed?
- A. 61.2 kJ
- B. 6.12 kJ
- C. 3.06 kJ
- D. 15.3 kJ
Q30. A maize plant (C₄) and a bean plant (C₃) are sealed together in a bright, transparent chamber. After several days, which outcome is most likely?
- A. Maize stops first because its PEP carboxylase has a low affinity for CO₂
- B. Both plants stop growing together as the CO₂ level falls to zero
- C. Bean keeps growing at a CO₂ level too low for the maize plant to gain carbon
- D. Maize keeps growing at a CO₂ level too low for the bean plant to gain carbon