NUMS — Medical & Dental Entry Test

NUMS Biology — Bioenergetics (Set 5)

Take the NUMS Biology — Bioenergetics (Set 5): 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. In the light reactions, which photosystem passes the high-energy electrons it receives from the chain on to NADP⁺ reductase to form NADPH?
- A. Photosystem I (P700) ✓
- B. Cytochrome b₆f complex
- C. Photosystem II (P680)
- D. ATP synthase complex
Answer: A. Photosystem I (P700)
Explanation: Photosystem I re-energises electrons that arrive from PSII and hands them via ferredoxin to NADP⁺ reductase.
Q2. Yeast ferments 90 g of glucose (molar mass 180 g/mol) completely to ethanol (46 g/mol) and CO₂. What mass of ethanol is formed?
- A. 92 g
- B. 23 g
- C. 46 g ✓
- D. 90 g
Answer: C. 46 g
Explanation: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. 90 g = 0.5 mol glucose, giving 1 mol ethanol = 46 g.
Q3. Fixing 12 molecules of CO₂ in the Calvin cycle needs 3 ATP and 2 NADPH per CO₂. What total supply of ATP and NADPH is needed?
- A. 36 ATP and 12 NADPH
- B. 18 ATP and 12 NADPH
- C. 24 ATP and 36 NADPH
- D. 36 ATP and 24 NADPH ✓
Answer: D. 36 ATP and 24 NADPH
Explanation: Per CO₂ fixed: 3 ATP and 2 NADPH; for 12 CO₂ that is 36 ATP and 24 NADPH.
Q4. On a hot afternoon, stomata close in both wheat (C3) and maize (C4) leaves, so internal CO₂ falls and O₂ rises. Which statement explains why maize still photosynthesises well?
- A. Maize fixes CO₂ with Rubisco in the mesophyll without any loss of ATP
- B. Rubisco in maize cannot bind O₂ and so avoids photorespiration
- C. Maize closes its stomata only at night and opens them in the afternoon
- D. PEP carboxylase keeps fixing CO₂ and does not react with O₂ ✓
Answer: D. PEP carboxylase keeps fixing CO₂ and does not react with O₂
Explanation: In C4 plants PEP carboxylase has a high affinity for CO₂ and no oxygenase activity, and the CO₂ is then concentrated near Rubisco in the bundle sheath, suppressing photorespiration.
Q5. A leaf sample is found to be very acidic at dawn and less acidic late in the afternoon. This pattern in a desert plant suggests
- A. CAM metabolism, with malic acid stored overnight and used by day ✓
- B. C3 metabolism, with citric acid stored by day and used overnight
- C. C4 metabolism, with malic acid made at dawn and used at dusk
- D. CAM metabolism, with malic acid made by day and used overnight
Answer: A. CAM metabolism, with malic acid stored overnight and used by day
Explanation: CAM plants open stomata at night, fix CO₂ into malic acid (stored in the vacuole, so sap is acidic by dawn), then decarboxylate it by day behind closed stomata, so acidity falls.

The other 25 questions

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Q6. A plant in the dark takes up 4 units of O₂ per hour. In moderate light, it releases 6 units of O₂ per hour (net). What is the rate of gross O₂ production by photosynthesis in that light?
- A. 10 units per hour
- B. 2 units per hour
- C. 4 units per hour
- D. 6 units per hour
Q7. At low light intensity, raising the CO₂ concentration from 0.04% to 0.10% does not change the rate of photosynthesis of a leaf, but raising the light intensity does. Which conclusion is correct?
- A. Water is the limiting factor under these conditions
- B. Light is the limiting factor under these conditions
- C. Temperature is the limiting factor under these conditions
- D. CO₂ is the limiting factor under these conditions
Q8. In paper chromatography of leaf pigments, a pigment spot lies 3.6 cm from the origin while the solvent front has moved 8.0 cm. What is its Rf value?
- A. 4.4
- B. 2.22
- C. 0.55
- D. 0.45
Q9. DCMU blocks electron flow from photosystem II. If a chloroplast with DCMU is illuminated, which process could still occur?
- A. Non-cyclic photophosphorylation making ATP and NADPH
- B. Cyclic photophosphorylation by photosystem I making ATP
- C. Photolysis of water releasing oxygen and protons
- D. Reduction of NADP⁺ using electrons from water
Q10. Plants are supplied with water in which all oxygen atoms are normal ¹⁶O, but with CO₂ containing the heavy isotope ¹⁸O. In which products would ¹⁸O be found after photosynthesis?
- A. Oxygen gas and glucose
- B. Oxygen gas and water
- C. Glucose and water
- D. Oxygen gas only
Q11. In the presence of the uncoupler DNP, mitochondria consume O₂ rapidly but make little ATP, and the tissue becomes warmer. Which explanation is correct?
- A. Protons are pumped faster into the matrix, so more ATP synthase works
- B. Protons leak back across the inner membrane, so energy is released as heat
- C. Pyruvate cannot enter the mitochondrion, so the cycle stops
- D. Electron transport is blocked, so NADH builds up in the matrix
Q12. How many molecules of CO₂ are released in the Krebs cycle alone when one molecule of glucose is completely oxidised in a mitochondrion?
- A. 6
- B. 4
- C. 3
- D. 2
Q13. One glucose molecule is completely oxidised through glycolysis, the link reaction and the Krebs cycle. How many reduced coenzyme molecules (NADH + FADH₂) are formed in total?
- A. 12
- B. 24
- C. 10
- D. 8
Q14. In cyanide poisoning, cytochrome oxidase is inhibited. Why does the Krebs cycle also slow down?
- A. Citrate synthase is inhibited directly by cyanide
- B. Acetyl-CoA can no longer be made from fatty acids
- C. NAD⁺ and FAD are not regenerated from NADH and FADH₂
- D. Oxaloacetate is converted irreversibly to citrate
Q15. Mature mammalian red blood cells have no mitochondria, yet they make ATP. Which end product of glucose breakdown do they release?
- A. CO₂, formed from the Krebs cycle
- B. Lactate, formed from pyruvate
- C. Ethanol, formed from pyruvate
- D. Acetyl-CoA, formed from pyruvate
Q16. Yeast growing in an anaerobic fermenter is suddenly supplied with oxygen. Which changes follow?
- A. Ethanol production stays the same and CO₂ release stops
- B. Ethanol production falls and glucose use per unit of growth rises
- C. Ethanol production rises and glucose use per unit of growth rises
- D. Ethanol production falls and glucose use per unit of growth drops
Q17. A garden plant has pale yellow leaves and reduced photosynthesis. The soil lacks the metal ion that sits at the centre of the chlorophyll molecule. Which ion is it?
- A. Calcium
- B. Magnesium
- C. Manganese
- D. Iron
Q18. A plant is given ¹⁴CO₂ for two seconds and then killed. In maize (a C4 plant) the label first appears mainly in a four-carbon acid, while in sunflower (C3) it first appears in a three-carbon compound. Which statement is correct?
- A. Maize and sunflower both fix CO₂ first with RuBP, but at different rates
- B. Maize first fixes CO₂ with PEP to oxaloacetate, and sunflower with RuBP to PGA
- C. Maize first fixes CO₂ with RuBP to oxaloacetate, and sunflower with PEP to PGA
- D. Maize first fixes CO₂ with PEP to PGA, and sunflower with RuBP to oxaloacetate
Q19. In the leaf disc flotation method, discs sink after the air is removed and then rise as the assay proceeds. Discs kept in the dark in the same bicarbonate solution stay at the bottom. Which statement explains the difference?
- A. Respiration releases CO₂ that fills the air spaces and lifts the discs
- B. Respiration releases O₂ that refills the air spaces and lifts the discs
- C. Photosynthesis releases O₂ that refills the air spaces and lifts the discs
- D. Photosynthesis uses CO₂ and makes the discs lighter than the solution
Q20. During fermentation in yeast, which molecule acts as the final hydrogen acceptor that allows NAD⁺ to be regenerated?
- A. Acetaldehyde
- B. Pyruvate
- C. Oxygen
- D. Ethanol
Q21. After a 100 m sprint, an athlete keeps breathing heavily for several minutes. Which explanation fits best?
- A. Extra O₂ is needed to convert pyruvate into lactate in the muscles
- B. Extra CO₂ is needed to raise blood pH and slow the heart rate
- C. Extra O₂ is needed to clear lactate and restore ATP and creatine phosphate
- D. Extra O₂ is needed to make lactate from the glycogen of the muscle
Q22. In which step of glycolysis is ATP formed by substrate-level phosphorylation?
- A. 1,3-bisphosphoglycerate → 3-phosphoglycerate
- B. Glyceraldehyde-3-phosphate → 1,3-bisphosphoglycerate
- C. Fructose-6-phosphate → fructose-1,6-bisphosphate
- D. Glucose → glucose-6-phosphate
Q23. A cell is given fructose-1,6-bisphosphate instead of glucose. Through the remaining steps of glycolysis, what is the net ATP gain per molecule?
- A. 4 ATP
- B. 3 ATP
- C. 2 ATP
- D. 0 ATP
Q24. A mutant plant has thylakoids that make a normal proton gradient but lack functional ATP synthase. What is the most likely result in the light?
- A. H⁺ builds up inside thylakoids, and the Calvin cycle slows
- B. H⁺ builds up in the stroma, and the Calvin cycle speeds up
- C. H⁺ builds up inside thylakoids, and ATP synthesis rises
- D. H⁺ falls inside thylakoids, and the Calvin cycle slows
Q25. Compared with a sun plant, a shade plant typically has a lower light compensation point. What does this mean?
- A. It needs more light to match photosynthesis with respiration
- B. It reaches maximum photosynthesis at a higher light intensity
- C. It reaches zero net CO₂ exchange at a lower light intensity
- D. It respires faster than it photosynthesises at all light levels
Q26. A plant is grown under a lamp that gives only green light of high intensity. Compared with the same plant under white light of equal intensity, what happens?
- A. Photosynthesis is slower, because carotenoids block the green light
- B. Photosynthesis is slower, because chlorophylls absorb green light poorly
- C. Photosynthesis is faster, because chlorophylls absorb green light best
- D. Photosynthesis is the same, because all wavelengths drive the process equally
Q27. A leaf is kept at a very low light intensity. Raising its temperature from 15 °C to 25 °C causes almost no rise in the rate of photosynthesis. Which explanation is best?
- A. Light-dependent reactions are enzyme-controlled and denature at 25 °C
- B. Light-dependent reactions are limiting, and they are hardly temperature-sensitive
- C. The Calvin cycle is limiting, and it is hardly temperature-sensitive
- D. CO₂ is limiting, and its solubility falls sharply above 15 °C
Q28. In a mitochondrion actively making ATP, which statement about proton concentration is correct?
- A. The matrix has a higher H⁺ concentration than the intermembrane space
- B. The intermembrane space has a higher H⁺ concentration than the matrix
- C. The cytosol has a lower H⁺ concentration than the matrix
- D. The intermembrane space and the matrix have equal H⁺ concentrations
Q29. Germinating seeds in a sealed flask containing KOH solution cause a 18 mL fall in gas volume. In an identical flask with water instead of KOH, the volume falls by only 3 mL. What is the respiratory quotient?
- A. 0.17
- B. 0.83
- C. 0.60
- D. 1.20
Q30. Malic acid (C₄H₆O₅) is completely oxidised: C₄H₆O₅ + 3O₂ → 4CO₂ + 3H₂O. What is the respiratory quotient?
- A. 0.70
- B. 1.33
- C. 0.75
- D. 1.00